The first term of a series in A.P. is 17, the last term is <math><mrow><mo>−</mo><mn>1</mn><mn>2</mn><mfrac><mrow><mn>3</mn></mrow><mrow><mn>8</mn></mrow></mfrac></mrow></math> and the sum of the series is <math><mrow><mn>2</mn><mn>5</mn><mfrac><mrow><mn>7</mn></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac></mrow></math> Find the common difference.

Let n = the number of terms.Then, 25716=n2 [17+ (−1238)]=n2×458⇒ 40716=37 n16 ⇒ n=40737=11Let d be the common difference.Then −1238 (= the eleventh term) = 17 + 10d⇒ 10d=−1238−17=−2358⇒ d=− 2358×10=−4716

The first term of a series in A.P. is 17, the last term is&nbsp;<math><mrow><mo>−</mo><mn>1</mn><mn>2</mn><mfrac><mrow><mn>3</mn></mrow><mrow><mn>8</mn></mrow></mfrac></mrow></math>&nbsp;and the sum of the series is&nbsp;<math><mrow><mn>2</mn><mn>5</mn><mfrac><mrow><mn>7</mn></mrow><mrow><mn>1</mn><mn>6</mn></mrow></mfrac></mrow></math>&nbsp;Find the common difference.

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The first term of a series in A.P. is 17, the last term is $- 12 \frac{3}{8}$ and the sum of the series is $25 \frac{7}{16}$ Find the common difference.

Option 1 -

$\frac{- 23}{12}$

Option 2 -

$\frac{- 47}{16}$

Option 3 -

$\frac{- 53}{14}$

Option 4 -

$\frac{- 39}{18}$

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Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

Let n = the number of terms.

Then, $25 \frac{7}{16} = \frac{n}{2} [17 + (- 12 \frac{3}{8})] = \frac{n}{2} \times 4 \frac{5}{8}$

$\Rightarrow \frac{407}{16} = 37 \frac{n}{16}$ $\Rightarrow n = \frac{407}{37} = 11$

Let d be the common difference.

Then $- 12 \frac{3}{8}$ (= the eleventh term) = 17 + 10d

$\Rightarrow 10 d = - 12 \frac{3}{8} - 17 = - \frac{235}{8}$

$\Rightarrow d = \frac{- 235}{8 \times 10} = - \frac{47}{16}$

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The first term of a series in A.P. is 17, the last term is −1238 and the sum of the series is 25716 Find the common difference.

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The first term of a series in A.P. is 17, the last term is $- 12 \frac{3}{8}$ and the sum of the series is $25 \frac{7}{16}$ Find the common difference.