Question

There are 37 terms in an A.P ., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429 .write the A.P.​

Answer 1

Answer :

AP : 3, 7, 11, 15, 19, .... , 147

Formula applied :

• a$_{n}$ = a + (n - 1)d

Explanation :

Given,

No. of terms in AP = 37

Sum of middle 3 terms = 225

Middle term = (37 + 1)/2 = 19

→ a$_{18}$ + a$_{19}$ + a$_{20}$ = 225

→ (a + 17d) + (a + 18d) + (a + 19d) = 225

→ 3a + 54d = 225

→ 3(a + 18d) = 225

→ a + 18d = 75 ------ equation 1

Sum of last three terms = 429

→ a$_{35}$ + a$_{36}$ + a$_{37}$ = 429

→ a + 34d + a + 35d + a + 36d = 429

→ 3a + 105d = 429

→ 3(a + 35d) = 429

→ a + 35d = 143 ------- equation 2

Subtract eq 1 from 2

a + 35d = 143

a + 18d = 75

__________

17d = 68

___________

→ d = 68/17

$\boxed{ d = 4}$

Substitute d = 4 in equation 1

→ a + 18d = 75

→ a + 18(4) = 75

→ a = 75 - 7

$\boxed{a = 3}$

a$_{2}$ = a + d = 3 + 4 = 7

a$_{3}$ = a + 2d = 3 + 2(4) = 11

a$_{4}$ = a + 3d = 3 + 3(4) = 15

and so on..

a$_{37}$ = a + 36d = 3 + 36(4) = 147

Therefore,

AP : 3, 7, 11, 15, 19, .... , 147