find the sum of first 15 terms of a gp 2,4,8
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Answer:
Here a
1
=24,a
2
=21
Common difference d=a
2
−a
1
=21−24=−3
⇒a
n
=a
1
+(n−1)dn
th
term formula
We first find the value of n for which the value of n is zero
The next integer value of that n will be the term of first negative number
⇒a
n
=a
1
+(n−1)d
⇒0=24+(n−1)(−3)
⇒−24=(n−1)−3
⇒8=n−1
⇒n=9
Hence, 10
th
term will be the first negative term of given AP
a
10
=a
1
+(n−1)d=24+9×(−3)=−3
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