Question

if the sum of first 6 terms of an A.P is 36and that of the first 16 terms is 256 find the sum of first 10 terms

Answer 1

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Here's your answer...

Let the first term of the AP be a.
Let the common difference be d.
Then...
$s6 = \frac{6}{2} (2 a + (6 - 1)d) \\ 36= 3(2a + 5d) \\ 2a + 5d = 12$
And...
$s16 = \frac{16}{2} (2a + (16 - 1)d) \\ 256 = 8(2a + 15d) \\ 2a + 15d = 32$
Now, by solving the simultaneous equations...
$2a + 15d = 32 \\ 2a + 5d = 12$
_________________
$10d = 20 \\ d = 2$

2a + 10 = 12
2a = 2
a = 1

Now, we find the sum of the first ten terms.
$s10 = \frac{10}{2} (2 \times 1 + (10 - 1) \times 2) \\ = 5(2 + 18) \\ = 5 \times 20 = 100$
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