Question

Find the sum of 7 numbers in series 11,103,1005....

Answer 1

Type of sequences in which there is a common ratio between two successive terms like 2nd and 1st & 3rd and 2nd …. are called geometric sequence or geometric progression.

If a is the first term and r is the common ratio then
a, ar, ar²,.......ar^n-1

Sum of the n terms of a G.P

Sn= a(r^n - 1)/ (r-1)

Let S= 11+103+1005+……to n terms

S= (10+1)+(10²+3)+(10³+5)+.....to n terms

S= [10+10²+10³+......10^n] + [1+3+5+......to n terms]

S= S1+S2 = G.P + A.P

S1= a(r^n - 1)/ (r-1) &. S2= n/2[2a+(n-1)d]

Here, a= 10, r= 10 & a= 1,d=2

S1= [10(10^n-1/10-1)]

S1= [10(10^n -1/9)]

S1= 10/9(10^n-1)

S2= n/2[2a+(n-1)d]

S2= n/2[2×1+(n-1)2

S2= n/2[2+2(n-1)]

S2=( n/2) ×2 [ 1+(n-1)]

S2= n [n]= n²

S2= n²

S= S1+ S2

S= 10/9(10^n-1) +n²

Here n= 7 (given)

S7= 10/9 (10^7 -1) + 7²

S7=10/9( 10000000 -1) +49

S7= 10/9(9999999)+49

S7= 10 × 1111111 + 49

S7= 11111110 +49

S7= 11111159

Hence, the sum of 7 terms is 11111159.

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Hope this will help you....