Question

find the AP of first term -4 last term 29 and the sum of its term is 150 find the common difference AP

Answer 1

Step-by-step explanation:

a = -4

an = 29

S = 150 d = ?

sn=n/2(2a+n-1 d)

150 = 150

Answer 2

Answer

Let d be the common difference.

Given:

• first term =a=-4
• last term =l=29

Sum of all the terms,

$\sf \: S_n=150 \\ \\ \sf \: S_n= \frac{n}{2} [a+l] \\ \\ \sf \implies150= \frac{n}{2} [ - 4+29] \\ \\ \sf \implies \: 150 \times 2 = n \times 25 \\ \\\sf \implies \: 300 = n \times 25 \\ \\ \sf \implies \: n = \cancel \frac{300}{25} \\ \\ \sf \implies \: n=12$

There are 12 terms in total.

Therefore, 29 is the 10th term of the AP.

Now,

$\sf29=a+(12−1)d \\ \\ \sf \implies \: 29= - 4+11d \\ \\ \sf \implies \: 29 + 4=11d \\ \\ \sf \implies \:33 = 11d \\ \\ \sf \implies \: d = \cancel \frac{33}{11} \\ \\ \sf \implies \:d = 3$

The common difference is 3.