Skip to content
Question of 114

Q.Find the sum of the sequence 8,88,888,8888,…8, 88, 888, 8888, \ldots up to the nthn^{th} term. OR If AA and GG are the arithmetic mean and geometric mean respectively between two positive numbers, then prove that the numbers are A±(A+G)(A−G)A \pm \sqrt{(A+G)(A-G)}.

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2024Subjective· 4mImportance★★★★★
0% · 0/114 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Writing each term as 89(10k−1)\frac{8}{9}(10^k-1) turns the sum into a geometric series plus a linear term, giving Sn=881(10n+1−10−9n)S_n = \frac{8}{81}(10^{n+1}-10-9n).

Step 1. Sn=8+88+888+⋯S_n = 8 + 88 + 888 + \cdots to nn terms =8(1+11+111+⋯to n terms)= 8(1 + 11 + 111 + \cdots \text{to } n \text{ terms}).

Step 2. Each term like 111…(k ones)=10k−19111\ldots(k \text{ ones}) = \dfrac{10^k - 1}{9}. So Sn=89∑k=1n(10k−1)=89(∑k=1n10k−n)S_n = \dfrac{8}{9}\displaystyle\sum_{k=1}^{n}(10^k - 1) = \dfrac{8}{9}\left(\sum_{k=1}^{n}10^k - n\right).

Step 3. The geometric sum ∑k=1n10k=10(10n−1)10−1=10n+1−109\displaystyle\sum_{k=1}^{n}10^k = \dfrac{10(10^n-1)}{10-1} = \dfrac{10^{n+1}-10}{9}.

Step 4. So Sn=89(10n+1−109−n)=881(10n+1−10)−8n9=881(10n+1−10−9n)S_n = \dfrac{8}{9}\left(\dfrac{10^{n+1}-10}{9} - n\right) = \dfrac{8}{81}\left(10^{n+1}-10\right) - \dfrac{8n}{9} = \dfrac{8}{81}\left(10^{n+1} - 10 - 9n\right).

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.