Question

In an Ap, if 12th term is -13 and the sum of first 4 terms is 24, what is the sum of the first 10 terms?

Answer 1

Hope it will help you well

Answer 2

Sum of first 10 terms of given AP is 0.

Step-by-step explanation:

Given: $a_{12} = -13$ and $S_{4} = 24$

To find: $S_{10}$ .

Solution:

Since, we know that

$a_{n} = a+ (n-1) d$

So, $a_{12} = a+ 11d = -13$ ( Say (1) )

Also, we know that,

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

So,

$S_{4} = \frac{4}{2} [2a+(4-1) d] = 24\\\\\Rightarrow 2 [2a+ 3d] = 24 \\\\\Rightarrow 2a + 3d = 12$ ( Say (2) )

Now, we will solve equations (1) & (2) using elimination method.

So, multiply equation (1) with 2 and subtract equation (2) from it.

we get,

$2a + 22d - 2a - 3d = -26 -12\\\Rightarrow 19d= -38\\\Rightarrow d=-2$

Now, putting the value of 'd' in equation (1), we get,

$a+ 11(-2) =-13\\\Rightarrow a= 9$

So, with values a = 9 and d = -2, lets find $S_{10}$.

$\because S_{n} = \frac{n}{2} [2a+(n-1) d]\\\\\Rightarrow S_{10} = \frac{10}{2} [2\times 9+(10-1)(-2)]\\\\\Rightarrow S_{10} = 5 [18+(-18)] = 0$

So, Sum of first 10 terms of given AP is 0.