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Find the sum to $'n'$ terms of the series $10+84+734+\dots$

  1. $\frac{9(9^n+1)}{10} +1$
  2. $\frac{9(9^n-1)}{8} +1$
  3. $\frac{9(9^n-1)}{8} +n$
  4. $\frac{9(9^n-1)}{8} +n^2$
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5 Answers

Best answer
9 votes
9 votes

Ans). This is a arithmetico -geometric series . and the break down of each number in the sequence is as follows

$9{^1}+1, 9{^2}+3,9{^3}+5,......$

Now if we use substitution method we can find the right summation formula for the above sequence.

Take for example the first two terms of the sequence and add them we will get the sum as $10+84=94$

Now consider the option $D$ ,here we have $n=2,$ So ,we are trying to find the sum for first two numbers of the sequence.

$\dfrac{9(9^2-1)}{8}+2^2=94$. Same procedure can used to find the sum of first three numbers in the sequence

The answer is option $D$

edited by
20 votes
20 votes

10 + 84 + 734 +...= (9+ 1) + (9+ 3) + (9+ 5) +...

 Such a series is called AGP (Arithmetic Geometric Progression) . For solving such progressions we need to find

$\sum (GP+AP)$

  =a(rn-1)/(r-1) + n(2*a+(n-1)*d)/2

  =9(9n-1)/8 + n(2+2n-2)/2

  =9(9n-1)/8 + n2

So, answer is (D) 

3 votes
3 votes

option D
we need to find sum of series up to 'n' terms..
for n=1 sum =10
for n=2 sum =10+84=94
for n=3 sum =10+84+734= 828
now put value of n in each formula find compare. like for n=2
option A:-  9(92 +1)/10 +1 = 74.8 (rejected)

option B:-  9(92 -1)/8 +1 = 90+1=91  (rejected)

option C :- 9(92 -1)/8 +n = 90+2=92 (rejected)

option D :- 9(9 2 - 1)/8 +n = 90+4= 94 (Accepted Answer)

0 votes
0 votes
The given expression is 10+84+734+----

= (9+1)+(9^2+3)+(9^3+5)+........

=(9+9^2+9^3+-------nterms) + (1+3+5+------- netrms)

=9(9^n-1)/(9-1) + n{ 2+(n-1)2}/2

=9(9^n-1)/8+n^2

which is the option D.
Answer:

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