Posted by **simranpreet** on Monday, December 3, 2012 at 6:16am.

the sum of first 8 terms of an arithmetic progression is 156.the ratio of its 12th term to its 68th term is 1:5.calculate the first term and the fifteenth term.

please help this questions please.

- maths -
**Reiny**, Monday, December 3, 2012 at 9:43am
sum of 8 terms is 156

(8/2)(2a + 7d) = 156

4(2a + 7d) = 156

2a + 7d = 39

(a+11d) : (a + 67d) = 1 : 5

(a+11d)/(a+67d) = 1/5

5a + 55d = a + 67d

4a = 12d

a = 3d

sub that into 2a+7d=39

6d + 7d = 39

13d=39

**d=3 , then in a = 3d ---> a = 9**

first term is 9

term 15 = a+14d = 9+14(3) = 51

check:

sum(8) = (8/2)(18 + 7(3)) = 4(39) = 156 , checks !

term12 = a+11d = 42

term 68 = a+67d = 210

and 42/210 = 1/5 , YEahh!

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