Question

the 8th term of an AP is half of its second term and the 11th term exceeds one third of its fourth term by 1 .find the 15th term

Answer 1

. Let First Term = a 

Common difference = d

According to First Condition,

=>a8=12a2=>a8=12a2 

=>a+(8−1)d=12[a+(2−1)d]=>a+(8−1)d=12[a+(2−1)d]

=>2a+14d=a+d=>2a+14d=a+d

=>a+13d=0       ......(1)=>a+13d=0       ......(1)

According to 2nd Condition

=>a11=(13×a4)+1=>a11=(13×a4)+1

=>a+(11−1)d=13[a+(4−1)d]+1=>a+(11−1)d=13[a+(4−1)d]+1

=>3a+30d=a+3d+3=>3a+30d=a+3d+3

=>2a+27d=3      .....(2)=>2a+27d=3      .....(2)

Multiply by (1) by 2, we get 

=>2a+26d=0       .......(3)=>2a+26d=0       .......(3)

Subtract (3) from (2), We get 

=>d=3=>d=3 , put vaue of d in(1), we get 

=>a+13×3=0=>a=−39=>a+13×3=0=>a=−39

Now, a15=a+(15−1)d=−39+42=3a15=a+(15−1)d=−3