Question

If the sum of first 7 terms of an A.P is 49 and the sum of first 17 terms is 289 find the sum of first 30 term
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Answer 1

Answer:-

Given:

Sum of first 7 terms of an AP = 49.

Sum of first 17 terms = 289

We know that,

Sum of first n terms of an AP – S(n) = n/2 * [ 2a + (n - 1)d ]

Hence,

→ 7/2 * [ 2a + (7 - 1)d ] = 49

→ 2a + 6d = 49 * 2/7

→ 2( a + 3d ) = 14

→ a + 3d = 14/2

→ a + 3d = 7 -- equation (1)

Similarly,

→ 17/2 * [ 2a + 16d ] = 289

→ 17/2 * 2 * (a + 8d) = 289

→ a + 8d = 289 * 2/17 * 1/2

→ a + 8d = 17 -- equation (2)

Subtract equation (1) from (2).

→ a + 8d - (a + 3d) = 17 - 7

→ a + 8d - a - 3d = 10

→ 5d = 10

→ d = 10/5

→ d = 2

Substitute the value of d in equation (1).

→ a + 3 * 2 = 7

→ a = 7 - 6

→ a = 1

Now,

Sum of first 30 terms = 30/2 * [ 2(1) + (30 - 1)(2) ]

→ S(30) = 15 (2 + 58)

→ S(30) = 15 * 60

→ S(30) = 900

Therefore, the sum of first 30 terms of the given AP.

Answer 2

Answer:

GivEn:

Sum of first 7 terms of an AP is 49.

Sum of first 17 terms of an AP is 289.

⠀⠀⠀⠀⠀⠀⠀

To find:

Sum of first n terms.

⠀⠀⠀⠀⠀⠀⠀

SoluTion:

⠀⠀⠀⠀⠀⠀⠀

⠀

${\underline{\sf{\bigstar\; According\;to\:question\;:}}} </p><p></p><p>$⠀⠀⠀⠀⠀

Sum of first 7 terms of an AP is 49.

⠀⠀⠀⠀⠀⠀⠀

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

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

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

⠀⠀⠀⠀⠀⠀⠀

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

$:\implies\sf 14 = 2a + 6d</p><p>$

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

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

⠀⠀⠀⠀⠀⠀⠀

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

Similarly,

⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 289 = \dfrac{17}{2}\bigg(2a + 16d \bigg)</p><p></p><p>⠀⠀⠀⠀⠀⠀⠀$

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

$:\implies\sf \cancel{289} \times \dfrac{2}{ \cancel{17}} </p><p>$

$:\implies\sf 34 = 2a + 16d</p><p>$

⠀⠀⠀⠀⠀⠀⠀

$</p><p>:\implies\sf 34 = 2(a + 8d)</p><p></p><p>$

$:\implies\sf \cancel{ \dfrac{34}{2}} = a + 8d</p><p>⠀⠀⠀⠀⠀⠀⠀$

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

━━━━━━━━━━━━━━⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

✒We get,

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 5d = 10</p><p>$

$:\implies\sf d = \cancel{ \dfrac{10}{5}}</p><p>$

$:\implies{\underline{\boxed{\sf{\pink{d = 2}}}}}\;\bigstar</p><p>$

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

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf 7 = a + 3 \times 2</p><p>$

$:\implies\sf 7 = a + 6</p><p></p><p>⠀⠀⠀⠀$⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

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

━━━━━━━━━━━━━━━

⠀⠀⠀⠀⠀⠀⠀

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

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

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

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

$:\implies\sf S_n = \dfrac{n}{2}\bigg( \cancel{2} + 2n \cancel{- 2} \bigg)</p><p>$⠀⠀⠀⠀⠀⠀⠀

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

⠀⠀⠀⠀⠀⠀⠀

$</p><p>:\implies\sf S_n = \dfrac{n}{ \cancel{2}} \times \cancel{2}$

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

∴ Sum of first n terms of an AP is n².