Question

If the first term of the AP is 4 and the sum of the first five terms is equal to one-fourth of the sum of the
next five terms, then find the 15th term?​

Answer 1

$\huge {\underline {\underline {\mathbb {\green {Answer:-}}}}}$

given

1st term or a = 4

6th term = a+5d

sum of n terms in an a.p = $\frac {n}{2} (2a+(n-1)d)$

so sum of first 5 terms of an a.p = $\frac {5}{2}(2×4+(4)d)$

and sum of next five terms of a.p = $\frac {5}{2}(2(a+5d)+4d)$

given

sum of first five terms = $\frac {1}{4}$sum of next five terms

=> $\frac {5}{2}(8+4d) = \frac {1}{4}(\frac {5}{2}(2a+14d))$

=>4(8+4d) = 8+14d

=>32+16d = 8+14d

=>16d-14d=8-32

=>2d= -24

=>d= -24/2= -12

let the 15th term of the sereis be z

z = a+14d

z = 4+14(-12)

z = 4-168= -164

$\huge so \:15th\: term = -164$

hope it helps

mark as BRAINLIEST

: )

Answer 2

Answer:

Step-by-step explanation:

1st term or a = 4

6th term = a+5d

sum of n terms in an a.p =  

so sum of first 5 terms of an a.p =  

and sum of next five terms of a.p =  

given

sum of first five terms = sum of next five terms

=>  

=>4(8+4d) = 8+14d

=>32+16d = 8+14d

=>16d-14d=8-32

=>2d= -24

=>d= -24/2= -12

let the 15th term of the sereis be z

z = a+14d

z = 4+14(-12)

z = 4-168= -164