Question

find the ap whose 3rd term is - 13 and 6th term is 2​

Answer 1

Given :

• In an AP 3rd term is -13 and 6th term is 2

To find :

• Series of that Ap

Formula used :

• Genral term of an AP is
• $\sf\:a_{n}=a+(n-1)d$

Solution :

• We have to find series of AP. Let the first term of an AP is a and common difference of series.

$\sf\:a_{3}=a+(3-1)d$

$\sf\:\implies\:a_{3}=a+2d$

$\sf\:\implies\:-13=a+2d....(1)$

and

$\sf\:a_{6}=a+(6-1)d$

$\sf\:\implies\:a_{6}=a+5d$

$\sf\:\implies\:2=a+5d...(2)$

Now solve, Equations (1)&(2)

Subratct equation (1) from (2)

$\implies \sf \:a + 5d - a - 2d = 2 + 13$

$\sf \implies3d = 15$

$\sf \implies \: d = \cancel{ \frac{15}{3}} = 5...(3)$

Now put the value of d in Equation (1)

$\sf \implies \: a + 2(5) = - 13$

$\sf \: \implies \: a = - 13 - 10 = - 23$

Now Ap series

First term = a = -23

second term =a+d = -23+5= -18

third term = a+2d = -23+10= -13

Therefore,AP series :

-23,-18,-13......

_________________________

Verification:

3rd term is -13

• Third term = a+2d = -13

Sixth term is 2

• Sixth term = a+5d = -23+25 = 2

Hence verified!