# Arithmetic and Geometric Progressions

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Infinite terms of a series

## Sum of infinite terms

If ${S}_{n} = {n}^{3} + {n}^{2} + n + 1$ , where ${S}_{n}$ denotes the sum of the first n terms of a series and ${t}_{m}$ = 291, then m is equal to?
1. 24
2. 30
3. 26
4. 20

Choice B. 30

## Detailed Solution

The pattern is as follows : 2, 23, 234...

${S}_{n} - {S}_{n-1} = {t}_{n}$

Substitute m instead of n
${S}_{m} - {S}_{m-1} = {t}_{m}$
We know that ${S}_{n} = {n}^{3} + {n}^{2} + n + 1$
Hence ${m}^{3} + {m}^{2} + m + 1$
${m}^{3} + {m}^{2} + m + 1 - [{(m-1)}^{3} + {(m-1)}^{2} + (m-1) + 1 ] = 291$
${m}^{3} + {m}^{2} + m + 1 - [{(m)}^{3} - {(3m)}^{2} + 3m - 1 + {(m)}^{2} - 2m + 1 + m - 1 + 1] = 291$
$1 + {3m}^{2} + 3m + 1 -2m - 1 = 291$
${-3m}^{2} + m - 290 = 0$
${3m}^{2} - m + 290 = 0$
Solving above equation we get m = -29, 30
M cannot be negative
Hence m = 30

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## More questions from Progressions

With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. Reinforce these ideas with these questions.