Question

If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first

n terms of the A.P​

Answer 1

GivEn:

• Sum of first 7 terms of an AP is 49.

• Sum of first 17 terms of an AP is 289.

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To find:

• Sum of first n terms.

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SoluTion:

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${\underline{\sf{\bigstar\; According\;to\:question\;:}}}$

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Sum of first 7 terms of an AP is 49.

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$:\implies\sf S_7 = \dfrac{7}{2}\bigg(2a + (7 - 1)d \bigg)$

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$:\implies\sf 49 = \dfrac{7}{2}\bigg(2a + 6d \bigg)$

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$:\implies\sf 49 \times \dfrac{2}{7} = 2a + 6d$

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$:\implies\sf \cancel{49} \times \dfrac{2}{ \cancel{7}} = 2a + 6d$

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$:\implies\sf 14 = 2a + 6d$

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$:\implies\sf 14 = 2(a + 3d)$

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$:\implies\sf \cancel{ \dfrac{14}{2}} = a + 3d$

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$:\implies\sf 7 = a + 3d\;\;\;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq\;(1)\bigg\rgroup$

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Similarly,

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$:\implies\sf S_{17} = \dfrac{17}{2}\bigg(2a + (17 - 1)d \bigg)$

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$:\implies\sf 289 = \dfrac{17}{2}\bigg(2a + 16d \bigg)$

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$:\implies\sf 289 \times \dfrac{2}{17} = 2a + 16d$

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$:\implies\sf \cancel{289} \times \dfrac{2}{ \cancel{17}} = 2a + 16d$

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$:\implies\sf 34 = 2a + 16d$

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$:\implies\sf 34 = 2(a + 8d)$

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$:\implies\sf \cancel{ \dfrac{34}{2}} = a + 8d$

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$:\implies\sf 17 = a + 8d\;\;\;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq\;(2)\bigg\rgroup$

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${\underline{\sf{\bigstar\;Now,\;Substracting\;eq(1)\;from\;eq(2)\;:}}}$

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we get,

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$:\implies\sf 5d = 10$

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$:\implies\sf d = \cancel{ \dfrac{10}{5}}$

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$:\implies{\underline{\boxed{\sf{\pink{d = 2}}}}}\;\bigstar$

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${\underline{\sf{\bigstar\;Now,\;Putting\;value\;of\;d\;in\;eq(1)\;:}}}$

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$:\implies\sf 7 = a + 3 \times 2$

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$:\implies\sf 7 = a + 6$

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$:\implies\sf a = 7 - 6$

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$:\implies{\underline{\boxed{\sf{\blue{a = 1}}}}}\;\bigstar$

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${\underline{\sf{\bigstar\;Sum\;of\;first\;n\;terms\;of\:AP\;is\;:}}}$

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$:\implies\sf S_n = \dfrac{n}{2}\bigg(2a + (n - 1)d\bigg)$

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$:\implies\sf S_n = \dfrac{n}{2}\bigg(2 \times 1 + (n - 1)2 \bigg)$

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$:\implies\sf S_n = \dfrac{n}{2}\bigg(2 + 2n - 2 \bigg)$

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$:\implies\sf S_n = \dfrac{n}{2}\bigg( \cancel{2} + 2n \cancel{- 2} \bigg)$

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$:\implies\sf S_n = \dfrac{n}{2} \times 2n$

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$:\implies\sf S_n = \dfrac{n}{ \cancel{2}} \times \cancel{2}n$

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$:\implies{\underline{\boxed{\sf{\purple{S_n = n^2}}}}}\;\bigstar$

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$\therefore$ Sum of first n terms of an AP is n².