Question

If the sum of the first 7 terms of an A.P is 49 and that of first 17 terms is 289 , find the sum of first 30 terms​

Answer 1

given:

S7 =49

S17=289

Step-by-step explanation:

S7=7/2×2a+(n-1)d

49=

Answer 2

Step-by-step explanation:

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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⠀

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

Sum of first 7 terms of an AP is 49.

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$:\implies\sf S_7 = \dfrac{7}{2}\bigg(2a + (7 - 1)d \bigg):⟹S </p><p>7</p><p> </p><p> = </p><p>2</p><p>7</p><p>$

$(2a+(7−1)d) \\ </p><p></p><p>:\implies\sf 49 = \dfrac{7}{2}\bigg(2a + 6d \bigg):</p><p>⟹49= </p><p>2</p><p> </p><p>7$

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

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$49</p><p> × </p><p>7</p><p> </p><p> </p><p>2</p><p> </p><p> =2a+6d</p><p></p><p>:\implies\sf 14 = 2a + 6d \\ :⟹14=2a+6d$

$:\implies\sf \cancel{49} \times \dfrac{2}{ \cancel{7}} = 2a + 6d \\ :⟹ </p><p></p><p></p><p>:\implies\sf 14 = 2(a + 3d): \\ ⟹14=2(a+3d) \\ </p><p></p><p>:\implies\sf \cancel{ \dfrac{14}{2}} = a + 3d:⟹ </p><p>2</p><p>14</p><p> </p><p> </p><p> </p><p> =a+3d$

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$:\implies\sf 7 = a + 3d\;\;\;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq\;(1)\bigg\rgroup:⟹7=a+3d </p><p>⎩</p><p> </p><p> eq(1)$

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Similarly,

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$:\implies\sf S_{17} = \dfrac{17}{2}\bigg(2a + (17 - 1)d \bigg): \\ ⟹S </p><p>17</p><p> </p><p> = </p><p>2</p><p>17</p><p> </p><p> (2a+(17−1)d)</p><p>$

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

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

$:\implies\sf \cancel{289} \times \dfrac{2}{ \cancel{17}}:⟹ </p><p>289</p><p> × </p><p>17</p><p> </p><p>2</p><p> </p><p> </p><p></p><p> \\ :\implies\sf 34 = 2a + 16d: \\ ⟹34=2a+16d</p><p>$

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$:\implies\sf 34 = 2(a + 8d): \\ ⟹34=2(a+8d)</p><p></p><p>:\implies\sf \cancel { \dfrac{34}{2}} = a + 8d ⠀⠀⠀⠀⠀⠀⠀ :⟹ </p><p>2</p><p>34 </p><p> </p><p> </p><p> </p><p> =a+8d⠀⠀⠀⠀⠀ \\ ⠀⠀</p><p></p><p>:\implies\sf 17 = a + 8d\;\;\;\;\;\;\;\;\;\;\;\;\bigg\lgroup\bf eq\;(2)\bigg\rgroup:⟹17=a+8d$

━

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

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✒We get,

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$:\implies\sf 5d = 10: \\ ⟹5d=10</p><p></p><p>:\implies\sf \\ d = \cancel{ \dfrac{10}{5}}:⟹d= </p><p>5</p><p>10 \\ </p><p> </p><p> </p><p> </p><p> </p><p></p><p>:\implies{\underline{\boxed{\sf{\pink{d = 2}}}}}\;\bigstar</p><p>$

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

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$:\implies\sf 7 = a + 3 \⟹7=a+3×2</p><p></p><p>:\implies\\ :⟹7=a+6⠀⠀ \\ ⠀⠀ ⠀⠀⠀</p><p></p><p>:\implies\sf a = 7 - 6</p><p></p><p>⠀⠀⠀⠀⠀⠀⠀$

$:\implies{\underline{\boxed{\sf{\blue{a = 1}}}}}\;\bigstar</p><p>$

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

$:\implies\sf S_n = \dfrac{n}{2}\bigg(2a + (n - 1)d\bigg): \\ ⟹S </p><p>n</p><p> </p><p> = </p><p>2</p><p>n</p><p> </p><p> (2a+(n−1)d)</p><p>$

$:\implies\sf S_n = \dfrac{n}{2}\bigg(2 \times 1 + (n - 1)2 \bigg): \\ ⟹S </p><p>n</p><p> </p><p> = </p><p>2</p><p>n</p><p> </p><p> (2×1+(n−1)2)$

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

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

$:\implies\sf S_n = \dfrac{n}{2} \times {2}: \\ ⟹S </p><p>n</p><p> </p><p> = </p><p>2</p><p>n</p><p> </p><p> ×2</p><p></p><p>⠀⠀⠀⠀⠀⠀⠀</p><p>$

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

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

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