Question

The sum of the first 30 terms of an AP is equal to the sum of its first 20 terms. show that sum of the 50 term is 0.​

Answer 1

$\huge\text{\underline{Answer}}$

$\sf{\underline{Given }}$

$\bold{S 20 = S30 }$

$\sf{\underline{To show }}$

$\bold{S50 = 0 }$

$\huge\sf{solution:}$

Let a be the first term and d be the common difference.

then, Sum of n th term in AP is given by :-

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

$\implies$ $\bold{ \frac{20}{2} [2a + (20 - 1)d] = \frac{30}{2} [2a + (30 - 1) d]}$

$\implies$ $\bold{2(2a + 19d) = 3(2a + 19d)}$

$\implies$ $\bold{4a + 38d = 6a + 87d }$

$\implies$ $\bold{2a + 49d = 0 }$ ----------eq. 1.

Now, sum of 50 th term of an AP.

$\implies$ $\bold{S50=\frac{50}{2}[ 2a + (50 - 1)d] }$

$\implies$ $\bold{S50 = [2a + 49d] }$

Putting the value of equation 1.

$\implies$ $\bold{= 25 × 0}$

$\implies$ $\bold{= 0}$

hence, sum of 50 term of AP is 0.

Answer 2

ANSWER:

0

STEP BY STEP EXPLANATION:

given,sum of first 30 terms=sum of first 20 terms

• S30=S20

->(30\2)[2a +30 -1 *d]=(20/2)[2a+(20-1)*d]

->15[2a+29d]=10[2a+19d]

->30a+435d=20a+190d

->10a=-245d

->a=24.5d

now,

sum of first 50 terms

=(50/2)[2a+50-1*d]

=25[49d+49*d]

=25[0]

=0

therefore ,sum of first 50 terms is zero

hope it helps

plz.. mark it as brainlist answer...