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Q.Find the sum to nn terms of the sequence 8,88,888,8888,…8, 88, 888, 8888, \ldots OR Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th4^{th} by 18.

Nagaland NbseNagaland Board of School Education (Class XI) 2021Subjective· 6mImportance★★★★★
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Express each term 8,88,888,…8,88,888,\ldots as 89(10k−1)\dfrac89(10^k-1) and sum a geometric series plus a constant. (Alternative: set up two equations for a 4-term G.P. from the given conditions and solve for aa and rr.)

Sum to nn terms of 8,88,888,8888,…8, 88, 888, 8888,\ldots:

The kk-th term is a string of kk eights: 88⋯8⏟k=8×11⋯1⏟k\underbrace{88\cdots8}_{k} = 8\times\underbrace{11\cdots1}_{k}.

A repunit of kk ones can be written as 11⋯1⏟k=10k−19\underbrace{11\cdots1}_{k} = \dfrac{10^k-1}{9} (e.g. 111=9999111 = \dfrac{999}{9}).

So the kk-th term is:

ak=8⋅10k−19=89(10k−1)a_k = 8\cdot\frac{10^k-1}{9} = \frac{8}{9}\left(10^k-1\right)

Sum to nn terms:

Sn=∑k=1nak=89∑k=1n(10k−1)=89[∑k=1n10k−n]S_n = \sum_{k=1}^{n} a_k = \frac{8}{9}\sum_{k=1}^{n}\left(10^k-1\right) = \frac{8}{9}\left[\sum_{k=1}^n 10^k - n\right]

The geometric sum ∑k=1n10k=10+102+⋯+10n=10(10n−1)10−1=10n+1−109\sum_{k=1}^n 10^k = 10+10^2+\cdots+10^n = \dfrac{10(10^n-1)}{10-1} = \dfrac{10^{n+1}-10}{9}.

So:

Sn=89[10n+1−109−n]=881(10n+1−10)−8n9=881(10n+1−9n−10)S_n = \frac{8}{9}\left[\frac{10^{n+1}-10}{9} - n\right] = \frac{8}{81}\left(10^{n+1}-10\right) - \frac{8n}{9} = \frac{8}{81}\left(10^{n+1}-9n-10\right)

(Check n=1n=1: S1=881(100−9−10)=881×81=8S_1=\frac{8}{81}(100-9-10)=\frac{8}{81}\times81=8 ✓. Check n=2n=2: S2=881(1000−18−10)=881×972=96=8+88S_2=\frac{8}{81}(1000-18-10)=\frac{8}{81}\times972=96=8+88 ✓.)

Or — Four numbers in G.P., third term exceeds first by 9, second exceeds fourth by 18:

Let the four numbers be a, ar, ar2, ar3a,\ ar,\ ar^2,\ ar^3.

"Third term is greater than the first term by 9": ar2−a=9  ⟹  a(r2−1)=9ar^2-a=9 \implies a(r^2-1)=9.

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