Question

An AP consist of 37terms. The sum of three middle most terms is 225 and the sum of the three last term is 429.Find the AP

Answer 1

hope this will help u......

Answer 2

Answer:

AP is 3,7,11,15,........

Step-by-step explanation:

An AP consist of 37 terms.

Middle most term = $\frac{n+1}{2}$

                              = $\frac{37+1}{2}$

                              = $\frac{38}{2}$

                              = $19$

So, 19 is the middlemost term

Now we are given that the sum of three middle most terms is 225

So, $a_{18}+a_{19}+a_{20} =225$

Formula of nth term: $a+(n-1)d$

So, $a+(18-1)d+a+(19-1)d+a+(20-1)d=225$

$a+17d+a+18d+a+19d=225$

$3a+54d=225$  ---A

Now we are given that  the sum of the three last term is 429

So, $a_{35}+a_{36}+a_{37} =429$

$a+(35-1)d+a+(36-1)d+a+(37-1)d=429$

$a+34d+a+35d+a+36d=429$

$3a+105d=429$  ---B

Subtract A from B

$3a+105d -3a-54d=429-225$

$51d=204$

$d=\frac{204}{51}$

$d=4$

Substitute the value of d in A

$3a+54(4)=225$

$3a+216=225$

$3a=9$

$a=3$

So, First term = 3

Second term = A+d=3+4=7

Third term = Second term +d =7+4=11

Fourth term = Third term +d = 11+4 =15

Hence AP is 3,7,11,15,........