Question

the sum of first five terms of an arithmetic sequence is 30 and the sum of first 7 terms is 56 what is the sum of its 6th and 7th term​

Answer 1

Given:

The sum of the first five terms of an arithmetic sequence is 30 and the sum of the first 7 terms is 56

To find:

The sum of its 6th and 7th term​

Solution:

The formula of the sum of n terms of an A.P. is:

$\boxed{\bold{S_n = \frac{n}{2} [2a + (n-1)d]}}$

where Sₙ = sum of n terms, a = first term, d = common difference and n = no. of terms

∴ $S _ 5 = \frac{5}{2} [2a + (5-1)d]}} = 30$

$\implies \frac{5}{2} [2a + 4d]}} = 30$

$\implies 5 [a + 2d] = 30$

$\implies [a + 2d] = 6$ . . . . (1)

∴ $S _ 7 = \frac{7}{2} [2a + (7-1)d]}} = 56$

$\implies \frac{7}{2} [2a + 6d]}} = 56$

$\implies 7 [a + 3d] = 56$

$\implies [a + 3d] = 8$ . . . . (2)

On subtracting equation (1) and (2), we get

a + 3d = 8

a + 2d = 6

-  -          -

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  d = 2

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On substituting the value of d in equation (1), we get

a + (2 × 2) = 6

⇒ a + 4 = 6

⇒ a = 2

We know,

$\boxed{\bold{nth\:term\:of\:an\:A.P., T_n = a + (n-1)d}}$

Therefore,

$T_6 = 2 + (6-1)2 = 2 + (5 \times 2) = 2 + 10 = 12$

and

$T_7 = 2 + (7-1)2 = 2 + (6 \times 2) = 2 + 12 = 14$

Now,

The sum of the 6th and 7th term is,

= $T_6 + T_7$

= 12 + 14

= 26

Thus, the sum of its 6th and 7th term​ is → 26.

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