Question

If 12th term of an A.P. is 13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Answer 1

Therefore the sum of first 10 terms is $\frac{1560}{19}$

Step-by-step explanation:

Given that the 12th term of an A.P is 13 and the sum of first four term is 24.

Let the the first term of the A.P be a.

And the common difference be d.

The nth term of an A.P is

$T_n=a+(n-1)d$

Then the 12th term of the A.P is

$T_{12}=a+(12-1)d$

$\Rightarrow 13= a+11d$ .....(1)      $[\because T_{12}=13]$

The sum of n term of an A.P is

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

Then the sum of four term is

$S_4=\frac{4}{2}[2a+(4-1)d]$

$\Rightarrow 24= 2(2a+3d)$

$\Rightarrow 12=2a+3d$.........(2)

Subtract 2 times of equation (1) from equation (2)

2a+3d-(2a+22d)=12-26

⇒2a +3d -2a -22d = -14

⇒ - 19d=-14

$\Rightarrow d= \frac{14}{19}$

Putting the value of d in equation (1)

$a+11\times \frac{14}{19}=13$

$\Rightarrow a= 13 - \frac{154}{19}$

$\Rightarrow a = \frac{93}{19}$

Therefore the sum of first 10 terms is

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

$=\frac{10}{2}[2\times \frac{93}{19}+(10-1)\frac{14}{19}]$    [n=10, $a = \frac{93}{19}$, $d= \frac{14}{19}$  ]

$=\frac{1560}{19}$