# Arithmetic and Geometric Progressions

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AP and GP

## Sum of all Terms

Find the sum of all the terms, If the first 3 terms among 4 positive integers are in A.P and the last 3 terms are in G.P. Moreover the difference between the first and last term is 40.
1. 108
2. 172
3. 124
4. 196

Choice B. 172

## Detailed Solution

Let the numbers be x-a, x, x+a

∴ For last 3 terms to be in G.P:-

(x+a)2 = x(x-a+40)

x2+a2+2ax = x2-ax+40x

a2 = -3ax + 40x

=) x = $\frac{a^2}{(40-3a)}$

Now, since x is a positive integer, putting different integral value of a .

Only a = 12 gives a integer value of x = 36

(Note, a=b also give integral value x=b but then the first term becomes 0)

∴ x = 36, a = 12

∴ x – a = 24

x = 36

x+a = 48

x-a+40 = 64

∴ Sum of all the terms = 64+48+36+24 = 172

The terms are - 24,36,48,64

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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.