Question

In an A.P .the sum of six terms is 147 and sum of nine terms is 315.hence find the series and also find sum of n terms

Answer 1

Step-by-step explanation:

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Answer 2

The series is 7, 14, 21, ......... and the sum of n terms is $\frac{n}{2}[7+7n]$.

Step-by-step explanation:

Let the first term of an A.P = a

Common difference = d

Sum of six terms = 147,    n = 6

$\frac{n}{2}[2a +(n-1)d] = 147\\=> \frac{6}{2}[2a +(6-1)d]=147\\=> 3[2a +5d]=147\\=>2a+5d=\frac{147}{3}=49..................(i)$

Sum of nine terms = 315, n = 9

$\frac{n}{2}[2a+(n-1)d]=315\\=>\frac{9}{2}[2a+(9-1)d]=315\\=>\frac{9}{2}[2a+8d]=315\\=>\frac{9}{2}\times 2[a+4d]=315\\=>9[a+4d]=315\\=>a+4d=\frac{315}{9} =35................(ii)$

Now, equation(ii) x 2 - equation(i),

2a + 8d - (2a + 5d) = 70 - 49

=> 3d = 21

=> d = $\frac{21}{3}$ = 7

Plug the value of d in equation(ii),

=> a + 4 x 7 = 35

=> a = 35 - 28 = 7

The series is a, a + d, a + 2d,................ = 7, 14, 21,.......

Sum of n terms =

$\frac{n}{2}[2a+(n-1)d] = \frac{n}{2}[2\times 7+(n-1)7] = \frac{n}{2}[7+7n]$

Hence, the series is 7, 14, 21, ......... and the sum of n terms is $\frac{n}{2}[7+7n]$.