Find the sum of the following series-

7 x 3 + 9 x 9 + 11 x 27 +13 x 81 + ……….. + up to 7 term

## Answer

Correct Answer : b ) 3^{10} – 6

Let S = 7 x 3 + 9 x 9 + 11 x 27 + 13 x 81 + ……….. + up to 7 term

Or, S = 7 x 3 + 9 x 3^{2} + 11 x 3^{3} + 13 x 3^{4} + ……….. + 19 x 3^{7} ……….. (1)

Now, Multiplying the above equation (1) by ‘3’ on both sides

3S = 7 x 3^{2} + 9 x 3^{3} + 11 x 3^{4} + ……….. + 19 x 3^{8} ………… (2)

Now subtracting equation (1) by equation (2) we get

-2S = 21 + 2 x 3^{2} + 2 x 3^{3} + ………. + 2 x 3^{7} – 19 x 3^{8}

-2S = 21 + 2 (3^{2} + 3^{3} + ………. + 3^{7}) – 19 x 3^{8}

We know that sum of a G.P. having common ratio (r > 1) will be given as

\(Sum=\frac{a(r^{n}-1)}{(r-1) }\)

Where a is the first term of the geometric progression (G.P.)

r is the common difference

and, n is the number of terms

Therefore,

-2S= 21 + 2(3^{2} (3^{n}-1)/(3-1)) – 19*3^{8}

-2S = 21 + 2(9(3^{6}-1)/(3-1)) – 19*3^{8}

-2S = 21 + 9(3^{6}-1)-19*3^{8}

-2S = 21 + 9*(3^{6} )-9-19*3^{8}

-2S = 21 - 9 + 9*(3^{6} ) - 19*3^{8}

-2S = 12 + (3^{8} ) - 19*3^{8}

-2S = 12 + (1-19)*3^{8}

-2S = 12 - 18*3^{8}

S= 9*3^{8 }- 6

S= 3^{10 }- 6

Hence, (b) is the correct answer.

Such type of question is asked in various government exams like SSC CGL, SSC MTS, SSC CPO, SSC CHSL, RRB JE, RRB NTPC, RRB GROUP D, RRB OFFICER SCALE-I, IBPS PO, IBPS SO, RRB Office Assistant, IBPS Clerk, RBI Assistant, IBPS RRB OFFICER SCALE 2&3, etc.

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