Question

if the sum of first 7 terms of an AP is 49 and that of 17 terms is 289 then the common difference of the AP is​

Answer 1

Given data,

The sum of the first 7 terms of an AP is 49 and the sum of the 17 terms is 289.

We know about the formula of the sum of n terms in Arithmetic progression that is $\red{\sf \: S_n = [2a + (n-1)d]}$

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Sum of first 7 terms = 49

$\sf\longmapsto S_n = \frac{n}{2} \big\lgroup2a + (n - 1)d \big\rgroup$

Putting values in formals

$\sf\longmapsto S_7 = \frac{7}{2} \big\lgroup2a + (n - 1)d \big\rgroup \\ \\ \sf\longmapsto 49= \frac{7}{2} \big\lgroup2a + (7 - 1)d \big\rgroup \\ \\ \ \sf\longmapsto 49= \frac{7}{2} \big\lgroup2a + (7 - 1)d \big\rgroup \\ \\ \ \sf\longmapsto 49= \frac{7}{2} \big\lgroup2a + 6d \big\rgroup \\ \\ \sf\longmapsto \frac{49 \times 2}{7} = 2a + 6d \\ \\ \sf\longmapsto \cancel\frac{98}{7} = 2a + 6d \\ \\ \sf\longmapsto14 = 2a + 6d \\ \\ \sf\longmapsto \frac{ \cancel{14} - \cancel6d}{ \cancel2} \:\:[dividing \;both\; number\; by \;2] \\ \\ \sf\longmapsto \green{a = 7 - 3d} - - - (1)$

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Sum of first 17 terms = 289

$\sf\longmapsto S_n = \frac{n}{2} \big\lgroup2a + (n - 1)d \big\rgroup$

Putting values in formals

$\sf\longmapsto S_{17} = \frac{17}{2} \big\lgroup2a + (17- 1)d \big\rgroup \\ \\ \sf\longmapsto 289 = \frac{17}{2} \big\lgroup2a + (17 - 1)d \big\rgroup \\ \\ \sf\longmapsto289= \frac{17}{2} \big\lgroup2a + 16d \big\rgroup \\ \\ \sf\longmapsto \frac{289 \times 2}{17} = \big\lgroup2a +16d \big\rgroup \\ \\ \sf\longmapsto \cancel\frac{578}{17} = \big\lgroup2a +16d \big\rgroup \\ \\ \sf\longmapsto34 = 2a + 16d \\ \\ \sf\longmapsto \frac{\cancel{34} - \cancel{16} d}{\cancel2} = a \: \: \: [dividing \;both\; number\; by \;2]\\ \\ \sf\longmapsto \red{a = 17 - 8d} - - - - (2)$

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We find the value of d From equation (1) and (2)

$\sf\longmapsto7 - 3d = 17 - 8d \\ \\ \sf\longmapsto8d - 3d = 17 - 7 \\ \\ \sf\longmapsto5d = 10 \\ \\ \sf\longmapsto \: d = \cancel \frac{10}{5} \\ \\ \sf\longmapsto \green{d = 2}$

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We find the value of a Putting values of d in equation (1)

$\sf\longmapsto \: a = 7 - 3d \\ \\ \sf\longmapsto \: a = 7 - 3 \times 2 \\ \\ \sf\longmapsto \: a = 7 - 6 \\ \\ \sf\longmapsto \green{a = 1}$

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We need to find sum of first n terms

We can use formula

$\sf\longmapsto S_n = \frac{n}{2} \big\lgroup2a + (n - 1)d \big\rgroup$

Putting values of a = 1 & d = 2

$\qquad \qquad \: \sf\longmapsto \: \frac{n}{2} \big\lgroup 2 \times 1 + \big(n - 1 \big)2\big\rgroup \\ \\ \qquad \qquad \: \sf\longmapsto \: \frac{n}{2} (2 + 2n - 2) \\ \\ \qquad \qquad \: \sf\longmapsto \: \frac{n}{2} (2 - 2 + 2n) \\ \\ \qquad \qquad \: \sf\longmapsto \: \frac{n}{2} (0 + 2n) \\ \\ \qquad \qquad \: \sf\longmapsto \: \frac{n}{ \cancel2} \times \cancel2n \\ \\ \qquad \qquad \: \sf\longmapsto \: \green{ {n}^{2} }$

The common difference of the AP be n²

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More information about AP

Supposition of an A.P.

• (1) = Three terms as: a- - d, a, a + d.

• (2) = Four terms as a - 3d, a - d, a +d, a + 3d.

• (3) = Five terms as a - 2d, a-d, a, a + d, a + 2d.