Question

12. If `1000 is deposited in a bank which pays annual interest at the rate of 5% compounded
annually, find the maturity amount at the end of 12 years .

Answer 1

GEOMETRIC PROGRESSION :
A sequence a1, a2, a3 ………….. an Is called a geometric sequence if a(n+1) = anr, where r is a non zero constant. Here, a1 is the first term and the constant r is called common ratio. The sequence is also called a geometric progression(G.P)

a1 = a , r = a(n+1)/ an

General term of a geometric sequence is
tn = arⁿ-1
General form of G.P is a, a r , a r ²,.........

SOLUTION IS IN THE ATTACHMENT
Here - Amount (A) = tn, Principal (P) = a
Rate( R) = r = (1 +R)

HOPE THIS WILL HELP YOU...

Answer 2

Solution :

Principal ( P ) = Rs 1000

Rate of interest ( r ) = 5%

Time ( T ) = 12 years

Number of times Interest paid

( n ) = 12

Let the amount = A

A = P( 1 + r/100 )ⁿ

=> A = 1000 [ 1 + 5/100 ]^12

=> A = 1000 ( 1 + 1/20 )^12

=> A = 1000 × ( 21/20 )^12

Therefore ,

Maturity amount at the end of

12 years = Rs 1000 × ( 21/20 )^12

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