Question

An

a.p consists of 12 terms. the sum of the two middle terms is 54 and the sum of last 3 terms is 135. find

a.p

Answer 1

n=12;
first equation,
(a+5d)+(a+6d)=54;

second equation,
(a+9d)+(a+10d)+(a+11d)=135;

on solving,we get,
27d=108;
(d=4;)
2a=10;
(a=5;)
so AP is,
5,(5+4),(9+4),(13+4)......
5,9,13,17.......

Hope answer helps you.....
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Answer 2

Answer:

The A.P is 55, 59, 63, 67.......

Step-by-step explanation:

Given an A.P consists of 12 terms. The sum of the two middle terms is 54 and the sum of last 3 terms is 135.

we have to find the A.P series

n=12;

As the sum of the two middle terms is 54

(a+5d)+(a+6d)=54

2a+11d=54 → (1)

and the sum of last 3 terms is 135.  

(a+9d)+(a+10d)+(a+11d)=135

3a+30d=135  → (2)

On solving, we get

27d=108;

d=4

2a+11(4)=154

2a+44=154 ⇒  a=55

so AP is,  

55, (55+4), (55+8), (55+12)......

55, 59, 63, 67.......