# Arithmetic and Geometric Progressions

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Investments and Progressions

## Total Amount

Ram invest different amounts during the year on shares. S1, S2, S3……….Sm are different sums of ‘n’ amounts invested in ‘m’ years. If the amounts invested during the years are in A.P whose first terms are 1,2,3…..m and common difference are 1,3,5…..,(2m-1) respectively then find the total amount invested by Ram in ‘m’ years.
1. n(m+1)
2. m+1
3. $\frac{mn}{2}$(mn+1)
4. cannot be determined

Choice C. $\frac{mn}{2}$(mn+1)

## Detailed Solution

Clearly, A/Q we have

S1 = $\frac{n}{2}$ * [2x1 + (n-1) x1] [∵ a=1 d=1]

S2 = $\frac{n}{2}$ * [2x2 + (n-1) x3] [∵ a=2 d=3]

S3 = $\frac{n}{2}$ * [2x3 + (n-1) x5] [∵ a=3 d=5]

Sm = $\frac{n}{2}$ * [2xm + (n-1) x (2m-1)] [∵a=m d= (2m-1)]

∴ (S1 + S2 + S3……+ Sm) = $\frac{n}{2}$ * [2x{1+2+3+4…+m} +(n-1)x{1+3+5….+(2m-1)}]

= $\frac{n}{2}$ * [{2x𝑚2(1+m)}+(n-1)x𝑚2{1+ (2m-1)}] [∵𝑆=𝑛2(𝑎+𝑙)]

= $\frac{n}{2}$ * [m(m+1) + m2(n-1)] = 𝑚𝑛2 [(m+1)+m(n-1)]

= $\frac{mn}{2}$(mn+1)

Correct Answer: $\frac{mn}{2}$ (mn+1)

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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.