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1. The fourth term of an A.P. is equal to 3 times the 1st term and the 7th term exceeds twice the 3rd term by 1. Find the 1st term and common difference?
2. Find three terms of an A.P. such that their sum is 3 and product is -8.
Ans1. T4 = 3T1
a+3d = 3a
3d = 2a
a = 3d/2 ....(1)
T7 = T3 +1
a+6d = a+2d +1
4d = 1
d=1/4
putting in (1)
we get , a=3/8...
Ans 2.Let the terms be a-d , a , a+d
Given that , a-d+a+a+d = 3
3a = 3
a=1
(a-d)(a)(a+d) = -8
(a2-d2)(a) =-8
since , a=1
(1-d2) = -8
d2 = 9
d=+-3
Now , the terms for a=1 & d=3 are : -2 , 1, 4
for a=1 & d=-3 are : 4 , 1 , -2
Hope this helps..
1.according to q.
a+3d=3a
3d-2a=0...................(1)
a+6d-(2{a+2d})=1
a+6d-(2a+4d)=1
-a+2d=1
a-2d=-1...................(2)
solving 1 and 2,we get,
a=3,d=2
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