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Question 5 - Finding sum from nth number - Chapter 8 Class 11 Sequences and Series
Last updated at July 28, 2026 by Teachoo
Chapter 8 Class 11 Sequences and Series
Concept wise
- Finding Sequences
- Arithmetic Progression (AP): Formulae based
- Arithmetic Progression (AP): Statement
- Arithmetic Progression (AP): Calculation based/Proofs
- Inserting AP between two numbers
- Arithmetic Mean (AM)
- Geometric Progression(GP): Formulae based
- Geometric Progression(GP): Statement
- Geometric Progression(GP): Calculation based/Proofs
- Inserting GP between two numbers
- Geometric Mean (GM)
- AM and GM (Arithmetic Mean And Geometric mean)
- AP and GP mix questions
- Finding sum from series
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Finding sum from nth number
Transcript
Question 5 Find the sum to n terms of the series 52 + 62 + 72 + .. + 202 52 + 62 + 72 + .. + 202 = (12 + 22 + 32 + 42 + 52 + 62 + + 202) (12 + 22 + 32 + 42) We know that Sum of square of n natural number is i.e. (12 + 22 + + n2) = (n(n+1)(2n+1))/6 For 12 + 22 + + 202 n = 20 Putting n = 20 in (2) 12 + 22 + + 202 = (20(20 + 1)(2(20) + 1))/6 = (20(21)(40 + 1))/6 = (20 (21) (41))/6 = (20 21 41)/6 = 10 7 41 = 2870 Thus, 12 + 22 + + 202 = 2870 For 12 + 22 + 32 + 42 n = 4 Putting n = 4 in (12 + 22 + + n2) = ( ( +1)(2 +1))/6 12 + 22 + 32 + 42 = (4(4 + 1)(2(4) + 1))/6 = (4 5 9)/6 = 2 5 3 = 30 Now, from (1) 52 + 62 + 72 + .. + 202 = (12 + 22 + 32 + 42 + 52 + 62 + + 202) (12 + 22 + 32 + 42) Putting values = 2870 30 = 2840 Thus, the required sum is 2840
