Question

17. The sum of the first and 8 terms and first 24 terms of an AP are equal. Find
the sum of its 32 terms.​

Answer 1

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Q- The sum of first 8 terms and first 24 terms of an A.P. are equal . Find the sum of first 32 terms .

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♡ Concept :

• In these type of Questions , we Simply derive relation between first term(a) and common Difference (d) .

• In this question , we will compare 8th and 24th term .

• Then we will get a relation between a and d .

• Then we will use that relation , to find the sum of first 32 terms .

♡ Given :

• $\rm \: s _{8} = s _{24}$
• It's an A.P.

♡ To Find :

• $\rm \: s _{32}$

♡ Solution :

• a = first term .
• d = common difference
• $\rm \: s _{n} = sum \: of \: n \: terms$

Since , we know that the sum of first 8 terms is equal to the sum of first 24 terms

so ,

equating them ..

$\rm \: s_{8} = s _{24}$

$\rm \mapsto \: \frac{ \cancel8}{ \cancel2} (2a + (8 - 1)d) = \frac{ \cancel{24}} {\cancel{2}} (2a + (24 - 1)d)$

$\mapsto \rm \: \cancel4(2a + 7d) = \cancel{ 12}(2a + 23d)$

$\mapsto \rm \: 2a + 7d = 3(2a + 23d)$

$\mapsto \rm \: 2a + 7d = 6a + 69d$

$\mapsto \rm \: 6a - 2a + 69d - 7d = 0$

$\mapsto { \rm \: 4a +62d = 0 }$

$\mapsto \boxed{ \rm \: 2a + 31d = 0}.....(i)$

Now ,

We have to find the sum of first 32 terms ..

so , using the Formula

$\rm \: s_{32} = \frac{32}{2} (2a + (32 - 1)d)$

$\mapsto \rm \: s _{32} = 16(2a + 31d)$

From eq(i) ,

• 2a + 31d = 0

$\mapsto \rm \: s _{32} = 16 \times 0$

$\large \boxed{ \rm \: s _{32} = 0}$

So , the sum of first 32 terms would be 0 .

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♡ Know More :

$1. \rm \: sum \: of \: n \: terms \: = \frac{n}{2} (2a + (n - 1)d)$

$2. \rm \: n \: th \: term \: = a + (n - 1)d$

Where ,

$\pink{ \ddot{ \smile}} \rm \: a = first \: term \:$

$\rm \blue{ \ddot{ \smile}} \: d = common \: difference \:$

$\green { \ddot{ \smile}} \rm \: n = {n}^{th} \: term \:$

Answer 2

Step-by-step explanation:

s 32 will be zero

hope it helps u

steps are mentioned above clearly