The 4th term of an A.P is equal to thrice the first term. The 7th term axceeds twice the third term by 1 then find-
1) first term
2) common difference
3) sum of first 10 terms
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Answer:
t4 = 3x t1
a+ 3d = 3(a)
a -3a +3d = 0
-2a +3d = 0 -------(1)
t7 = 2x t3 + 1
à +6d = 2(a +2d) +1
a +6d = 2a +4d +1
-a +2d = 1
-2a +4d = 2 -----(2)
-2a +3d = 0 -----(1)
+. -. -
______________( by subtracting)
d = -2
-2a +4d =2
-2a = 2+8.
a = -5
S10 = n/2 [ 2a + (n-1) d ]
= 5 [ -10 -18]
= -140
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