In an AP :
given S10 = 125, a3 = 15 find d and a10
d is the common difference of the series
a is the first term of the series
a3 = 15(Given in the Question)
We know that :-an= a + (n-1)d
∴ a3= a + (3-1)d
a3= a + 2d=15.......(i)
The sum of the series is Sn =
[2a + (n-1) d]∴S10 =10/2 [2×a + (10-1)×d]
⇒125 = 5[2a + 9d)⇒ 25 = 2a + 9d...... (ii)
Solving (i) and (ii) we get:-
2a + 9d= 25
a + 2d= 15 (Multiplying the Equation by 2 and changing the sign)
⇒ 2a + 9d = 25
2a + 4d = 30
- - -
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⇒ 5d = (-5)
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∴ d= (-1)
After putting value of d in any of the above equation we get a=17
∴ a 10 = 17 + (10-1)(-1)
= 8
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