{"id":2468,"date":"2021-10-01T04:54:54","date_gmt":"2021-10-01T04:54:54","guid":{"rendered":"http:\/\/stgwebsite.mindspark.in\/wordpress\/?page_id=2468"},"modified":"2022-01-02T06:32:22","modified_gmt":"2022-01-02T06:32:22","slug":"arithmetic-progression-definition-and-formulas","status":"publish","type":"page","link":"https:\/\/stgwebsite.mindspark.in\/studymaterial\/math-concepts\/arithmetic-progression-definition-and-formulas\/","title":{"rendered":"Arithmetic Progression- Definition and formulas"},"content":{"rendered":"

[et_pb_section fb_built=”1″ admin_label=”Section” module_class=”mainsec” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#e0f2fd” z_index=”1″ custom_padding=”5px||5px||true|false” locked=”off” collapsed=”off” global_colors_info=”{}”][et_pb_row column_structure=”3_5,2_5″ custom_padding_last_edited=”on|phone” _builder_version=”4.10.8″ _module_preset=”default” background_color=”#FFFFFF” width=”100%” max_width=”1310px” custom_padding=”|51px|40px|51px|false|true” custom_padding_tablet=”” custom_padding_phone=”|40px|30px|40px|false|true” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}”][et_pb_column type=”3_5″ admin_label=”Column L” _builder_version=”4.9.10″ _module_preset=”default” global_colors_info=”{}”][et_pb_text admin_label=”Acute Angles
\n” _builder_version=”4.10.8″ _module_preset=”default” header_font=”|700|||||||” header_text_align=”left” header_font_size=”50px” header_line_height=”1.18em” custom_padding=”|0px||4px|false|false” header_font_size_tablet=”” header_font_size_phone=”35px” header_font_size_last_edited=”on|phone” global_colors_info=”{}”]<\/p>\n

Arithmetic Progression- Definition and formulas<\/strong> <\/span><\/h1>\n

[\/et_pb_text][et_pb_text admin_label=”Text” _builder_version=”4.11.3″ _module_preset=”default” text_font_size=”16px” header_2_font=”|600|||||||” header_2_text_color=”#a01414″ header_3_font=”|600|||||||” header_3_text_color=”#898989″ custom_padding=”15px|15px||4px|false|false” global_colors_info=”{}”]<\/p>\n

Arithmetic Progression:<\/strong><\/h2>\n

Consider the series 1, 5, 9, 13, 17 or a series containing the terms 10, 16, 22, 28. Are you able to observe the pattern emerging in the above series? In both the examples given above, the difference between any two adjacent terms is the same. Such a series in which the common difference between any two adjacent terms is the same is known as an arithmetic progression.<\/p>\n

Let us learn various formulas and their derivations now.<\/p>\n

Formulas in Arithmetic Progression<\/strong><\/h2>\n

General Expression of arithmetic progression<\/strong><\/h3>\n

Let us start with the general form of an arithmetic progression. An AP can be expressed as a, a+d, a+2d, a+3d\u2026.. a+(n-1)d<\/p>\n

Where,<\/p>\n

a= first term<\/p>\n

d= common difference<\/p>\n

n= number of terms in the arithmetic progression.<\/p>\n

Nth term of an AP Series<\/strong><\/h3>\n

In an arithmetic progression where the first term and the common difference is known, the nth term of the series can be calculated using the formula Tn <\/sub>= a+(n-1)d<\/p>\n

 <\/p>\n

For example, we can calculate the 12th term of an AP having the first term a = 5 and the common difference d= 7 by substituting these values in the above formula and we get 82 as the 12th<\/sup> term of the series.<\/p>\n

Similarly, if we are to calculate the nth term from the end of the AP, then we use the formula
\nTn <\/sub>= l-(n-1)d<\/p>\n

Where<\/p>\n

l =\u00a0 the last term<\/p>\n

n = number of terms<\/p>\n

d = the common difference<\/p>\n

 <\/p>\n

Number of terms in an AP<\/strong><\/h3>\n

When the first and last terms of an AP and the common difference is known, then we can calculate the number of terms in the arithmetic progression using the formula<\/p>\n

n= ((l-a))\/d+1<\/p>\n

Where,<\/p>\n

l= the last term of the AP<\/p>\n

a= the first term of the AP<\/p>\n

d= the common difference<\/p>\n

n= number of terms in the AP<\/p>\n

Sum of a finite series<\/strong><\/h3>\n

When we know the first term and the common difference in the AP, the formula for calculating the sum of an arithmetic progression is S= (n[2a+(n-1)d])\/2<\/p>\n

 <\/p>\n

When we know the first and the last term, the sum of n terms in an AP is S= (n(a+l))\/2<\/p>\n

Where a is the first term and l is the last term.<\/p>\n

Derivation<\/strong><\/h2>\n

Let us say there are n terms in an arithmetic progression having the common difference d. The sum of such AP series would be <\/span>S=a + (a+d)+ (a+2d)+ \u2026.. [a+(n-1)d<\/span>]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span> (1)<\/span><\/p>\n

Writing the same in the reverse order would be\u00a0<\/span><\/p>\n

S= [a+(n-1)d] + [a+(n-2)d] + [a+(n-3)d]…. (a+d) + a<\/span> \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 <\/span> (2)<\/span><\/p>\n

Adding (1) and (2) we get,<\/span><\/p>\n

2S = [2a+(n-1)d] + [2a+(n-1)d] + [2a+(n-1)d]……[2a+(n-1)d] ( n times)<\/span><\/p>\n

I.e. <\/span>2S =n[2a+(n-1)d]<\/span><\/p>\n

I.e.<\/p>\n

The sum of first n terms S=(n[2a+(n-1)d])\/2..<\/p>\n

Formula summary<\/strong><\/h2>\n

 <\/p>\n\n\n\n\n\n\n\n
nth<\/sup> term of an AP<\/td>\nTn <\/sub>= a+(n-1)d<\/td>\n<\/tr>\n
nth<\/sup> term from the end of an AP<\/td>\nTn <\/sub>= l-(n-1)d<\/td>\n<\/tr>\n
Number of terms in an AP<\/td>\nn= ((l-a))\/d+1<\/td>\n<\/tr>\n
Sum of n terms in an AP<\/td>\n\u00a0S= (n[2a+(n-1)d])\/2<\/td>\n<\/tr>\n
Sum of n terms in an AP when first and last terms are known<\/td>\nS= (n(a+l))\/2<\/td>\n<\/tr>\n<\/thead>\n<\/table>\n

 <\/p>\n

Solved Examples<\/strong><\/h3>\n
    \n
  1. In an AP 5, 13, 21, 29, \u2026.. 261, find the following:\n
      \n
    1. Number of terms in the AP<\/li>\n
    2. 8th <\/sup>term from the beginning of the AP<\/li>\n
    3. 8th<\/sup> term from the end of the AP<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

      Solution:<\/strong><\/p>\n

      We can calculate the number of terms in an AP using the formula n= ((l-a))\/d+1<\/p>\n

      In the given question, the first term a = 5, last term l = 261, and the common difference d = 8.<\/p>\n

      Substituting these values, we get n= ((261-5))\/8+1<\/p>\n

      Thus, the number of terms in the AP is 33.<\/p>\n

      We know that we can calculate the nth term of an AP using the formula Tn<\/sub>=a+(n-1)d.<\/p>\n

      Substituting the values given in the above formula,<\/p>\n

      T8<\/sub>=5+(8-1)8<\/p>\n

      T8<\/sub>=61<\/p>\n

      Using the formula <\/span>T<\/span>n <\/span>= l-(n-1)d<\/span> we can calculate the 8th term from the end of the AP.<\/span><\/p>\n

      Substituting the figures from the question, we get <\/span>T<\/span>n <\/span>= 261-(8-1)8<\/span><\/p>\n

      Thus<\/span>, <\/span>T<\/span>n <\/span>= 205<\/span><\/p>\n

       <\/p>\n

      2. If the 8th and 10th terms of an AP are 57 and 71 respectively, find the 15th term of the AP\u00a0<\/span><\/p>\n

       <\/p>\n

      Solution:<\/b><\/p>\n

      We know that the nth term of an AP is <\/span>T<\/span>n <\/span>=a+(n-1)d<\/span><\/p>\n

      So, we get <\/span>T<\/span>8 <\/span>=57=a+(8-1)d<\/span><\/p>\n

      \u00a0<\/span>57=a+7d<\/span> (1)<\/span><\/p>\n

      Similarly, <\/span>T<\/span>10 <\/span>=71=a+(10-1)d<\/span><\/p>\n

      71=a+9d<\/span> (2)<\/span><\/p>\n

      Subtracting (1) from (2), we get,<\/span><\/p>\n

      14 = 2d.<\/span><\/p>\n

      d= 7<\/span><\/p>\n

      Substituting the value of d = 7 in (2), we get<\/span><\/p>\n

      71=a+9×7<\/span><\/p>\n

      a=8<\/b><\/p>\n

      Now,<\/span> T<\/span>12<\/span>=8+(12-1)7<\/span><\/p>\n

      T<\/b>12 <\/b>=85<\/b><\/p>\n

       <\/p>\n

      3. Find the sum of the first 20 terms of the AP 2, 7, 12, 17\u2026.<\/span><\/p>\n

      Solution:<\/strong><\/p>\n

      From the given question, we know that a=2 and d= ( 7-2)=5.<\/p>\n

      We know that the sum of first n numbers in an AP is S= (n[2a+(n-1)d])\/2<\/p>\n

      Substituting the values in the formula, we get<\/p>\n

      S20<\/sub>= (20[2\u27152+(20-1)\u27155])\/2<\/p>\n

      S20<\/sub>= (20[4+95])\/2<\/p>\n

      S20<\/sub>=[\/et_pb_text][et_pb_text disabled_on=”on|on|on” admin_label=”Sample Questions
      \n” _builder_version=”4.10.8″ _module_preset=”default” header_font=”|700|||||||” header_font_size=”28px” custom_padding=”|0px||4px|false|false” disabled=”on” global_colors_info=”{}”]<\/p>\n

      Practice Multiple Choice Questions<\/h1>\n

      [\/et_pb_text][et_pb_text disabled_on=”on|on|on” admin_label=”Question 1″ _builder_version=”4.10.8″ _module_preset=”default” background_color=”#FFFFFF” custom_padding=”25px|25px|25px|25px|true|true” border_radii=”on|15px|15px|15px|15px” border_width_all=”2px” border_color_all=”#000000″ border_width_top=”4px” border_color_top=”#E02B20″ disabled=”on” global_colors_info=”{}”]<\/p>\n

      \n
      \n

      Questions<\/strong><\/p>\n

        \n
      1. In an AP 5, 13, 21, 29, \u2026.. 261, find the following:<\/span>\n
          \n
        1. Number of terms in the AP<\/span><\/li>\n
        2. 8<\/span>th <\/span>term from the beginning of the AP<\/span><\/li>\n
        3. 8<\/span>th<\/span> term from the end of the AP<\/span><\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

          Solution:<\/b><\/p>\n

          We can calculate the number of terms in an AP using the formula <\/span>n=<\/span>(l-a)<\/span>d<\/span>+1<\/span><\/p>\n

          In the given question, the first term a = 5, last term l = 261, and the common difference d = 8.<\/span><\/p>\n

          Substituting these values, we get <\/span>n=<\/span>(261-5)<\/span>8<\/span>+1<\/span><\/p>\n

          Thus, the number of terms in the AP is 33.<\/span><\/p>\n

          We know that we can calculate the nth term of an AP using the formula <\/span>T<\/span>n <\/span>=a+(n-1)d<\/span>.<\/span><\/p>\n

          Substituting the values given in the above formula,\u00a0<\/span><\/p>\n

          T<\/span>8 <\/span>=5+(8-1)8<\/span><\/p>\n

          T<\/span>8 <\/span>=61<\/span><\/p>\n

          Using the formula <\/span>T<\/span>n <\/span>= l-(n-1)d<\/span> we can calculate the 8th term from the end of the AP.<\/span><\/p>\n

          Substituting the figures from the question, we get <\/span>T<\/span>n <\/span>= 261-(8-1)8<\/span><\/p>\n

          Thus<\/span>, <\/span>T<\/span>n <\/span>= 205<\/span><\/p>\n

            \n
          1. If the 8th and 10th terms of an AP are 57 and 71 respectively, find the 15th term of the AP\u00a0<\/span><\/li>\n<\/ol>\n

            Solution:<\/b><\/p>\n

            We know that the nth term of an AP is <\/span>T<\/span>n <\/span>=a+(n-1)d<\/span><\/p>\n

            So, we get <\/span>T<\/span>8 <\/span>=57=a+(8-1)d<\/span><\/p>\n

            \u00a0<\/span>57=a+7d<\/span> (1)<\/span><\/p>\n

            Similarly, <\/span>T<\/span>10 <\/span>=71=a+(10-1)d<\/span><\/p>\n

            71=a+9d<\/span> (2)<\/span><\/p>\n

            Subtracting (1) from (2), we get,<\/span><\/p>\n

            14 = 2d.<\/span><\/p>\n

            d= 7<\/span><\/p>\n

            Substituting the value of d = 7 in (2), we get<\/span><\/p>\n

            71=a+9×7<\/span><\/p>\n

            a=8<\/b><\/p>\n

            Now,<\/span> T<\/span>12<\/span>=8+(12-1)7<\/span><\/p>\n

            T<\/b>12 <\/b>=85<\/b><\/p>\n

              \n
            1. Find the sum of the first 20 terms of the AP 2, 7, 12, 17\u2026.<\/span><\/li>\n<\/ol>\n

              Solution:<\/b><\/p>\n

              From the given question, we know that a=2 and d= ( 7-2)=5.\u00a0<\/span><\/p>\n

              We know that the sum of first n numbers in an AP is S=<\/span> n[2a+(n-1)d]<\/span>2<\/span><\/p>\n

              Substituting the values in the formula, we get<\/span><\/p>\n

              S<\/span>20<\/span>=<\/span> 20[2\u27152+(20-1)\u27155]<\/span>2<\/span><\/p>\n

              S<\/span>20<\/span>=<\/span> 20[4+95]<\/span>2<\/span><\/p>\n

              S<\/span>20<\/span>=<\/span> 990<\/span><\/p>\n<\/div>\n<\/div>\n

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              Explore Other Topics<\/h1>\n

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              Geometry<\/a><\/div>\n
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              [\/et_pb_text][et_pb_text admin_label=”Related Concepts
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              Related Concepts<\/h1>\n

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              \n
              AP and GP<\/a><\/div>\n
              Sum of an infinite AP<\/a><\/div>\n
              {\\text{n}}^{\\text{th}}<\/span> term of an AP<\/a><\/a><\/div>\n<\/div>\n

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              Frequently Asked Questions\u00a0<\/span><\/span><\/h1>\n
                \n
              1. What is the general form of an arithmetic progression?<\/span><\/li>\n<\/ol>\n

                An Arithmetic Progression can be expressed in the form of a, a+d, a+2d\u2026. a+(n-1)d\u00a0<\/span>
                \nWhere,<\/span>
                \n\u00a0<\/span> a is the first term and<\/span>
                \nd is the common difference<\/span><\/p>\n

                2. What is the sum of first n numbers in an AP?<\/span>
                \nThe sum of first n numbers can be calculated using the formula S=<\/span> n[2a+(n-1)d]<\/span>2<\/span><\/p>\n

                3. What is the sum of first n natural numbers?<\/span>
                \nThe sum of first n natural numbers is<\/span> S= <\/span>n(n+1)<\/span>2<\/span>
                \nWhere n is the number of natural numbers.\u00a0<\/span><\/p>\n

                4. What is the formula to calculate the sum of squares of first n natural numbers?<\/span>
                \nWe can calculate the sum of squares of the first n natural numbers of an AP is\u00a0<\/span>
                \nS= <\/span>n(n+1)(2n+1)<\/span>6<\/span>[\/et_pb_text][\/et_pb_column][\/et_pb_row][\/et_pb_section]<\/p>\n","protected":false},"excerpt":{"rendered":"

                Formula for sum of terms in an AP S= n[2a+(n-1)d]2Formula for the nth term in an AP is a+(n-1)d where a = first term, d = difference, and n = number of terms.<\/p>\n","protected":false},"author":7,"featured_media":0,"parent":714,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_et_pb_use_builder":"on","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"yoast_head":"\nArithmetic Progression- Definition and formulas - mydomain<\/title>\n<meta name=\"description\" content=\"Formula for sum of terms in an AP S= (n[2a+(n-1)d])\/2Formula for the nth term in an AP is a+(n-1)d where a = first term, d = difference, and n = number of terms\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/stgwebsite.mindspark.in\/studymaterial\/math-concepts\/arithmetic-progression-definition-and-formulas\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Arithmetic Progression- Definition and formulas - 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