Question

In the following AP's find the missing terms:
(i)-4. __, __, __, __, 6
(ii) __, 38, __, __, __, -22

Answer 1

(i)

-4. __, __, __, __, 6

Here, First term = a = -4

$6^{th}$ term = $a_6$ = 6

Using formula $a_n$=a+ (n−1) d, to find nth term of arithmetic progression, we get

$a_6=-4+(6-1)d$

$\Rightarrow 6=-4+5d$

$\Rightarrow 5d=10$

$\Rightarrow$ $d=$ $\frac{10}{5}$ $=2$

Therefore, common difference = d = 2

Second term = first term + d = a+ d = -4 + 2 = -2

Third term = second term + d = -2 + 2 = 0

Fourth term = third term + d = 0 + 2 = 2

Fifth term = fourth term + d = 2 + 2 = 4

Therefore, missing terms are -2, 0, 2 and 4.

(ii)

__, 38, __, __, __, -22

In this problem, we are given 2nd and 6th term.

Using formula $a_n$=a + (n−1) d, to find nth term of arithmetic progression, we get

$a_2=a+(2-1)d$ and $a_6=a+(6-1)d$

$\Rightarrow 38=a+d$ and $-22=a+5d$

These are equations in two variables; we can solve them using any method. Let’s solve them using substitution method.

Using equation (38=a+ d), we can say that a=38−d. Putting value of a in equation (−22=a+5d), we get

$-22=38-d+5d$

$\Rightarrow 4d=-60$

$\Rightarrow$ d=− $\frac{60}{4}$=−15

Using this value of d and putting this in equation 38=a +d, we get

38=a−15

$\Rightarrow a=53$

Therefore, we get a=53 and d=−15

First term = a = 53

Third term = second term + d = 38 - 15 = 23

Fourth term = third term + d = 23 - 15 = 8

Fifth term = fourth term + d = 8 - 15 = -7

Therefore, missing terms are 53, 23, 8 and -7.