Question

Is the sum of 5th and 9th term of an ap is 723 and 7th and 12th term of an ap is 97 find AP which term of GP is 3, 9, 27 is 2187

Answer 1

Answer:

In an AP,

a5+a9=a+4d+a+8d

2a+12d=72———equation-1

a7+a12=a+6d+a+11d

2a+17d=97———equation-2

from eq-1 & eq-2

2a+12d=72

(–) 2a+17d=97

————————

-5d=-25

d=5

substitute d value in eq-1

2a+12(5)=72

2a+60=72

2a=72-60

a=12/2

a=6

a1=6

a2=a+d=6+5=11

a3=a2+d=11+5=16

therefore, the required AP:6,11,16............

In a GP,

a=3, r=9/3=3

let 2187 be nth term of GP

an=ar^n-1

2187=3(3)^n-1

2187=3¹×3^n-1. • • a^m×a^n=a^m+n

3^7=3^1+n-1

3ⁿ=3^7

→n=7

therefore, 2187 is 7 th term of the GP.

Hope it works......................................™✌️✌️