if for an A.P. s15 = 147 and s14= 123 find t15
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Answer:
147 = 15/2[ 2a + ( 15 - 1 ) d ]
147 = 15/2[ 2a + 14d ]
147×2/15 = [ 2a + 14d]
294/15 = 2a + 14d ---------(1)
123 = 14/2 [ 2a + ( 14 -1 ) d ]
123 × 7 = 2a + 13d
861 = 2a + 13d ------------------(2)
from (1) and (2)
294/15 = 2a + 14d
861 = 2a + 13d
-. -. -.
----------------------------
- 12621 = d
putting the value of d in (2)
861 = 2a + 13× - 12621
861 = 2a - 164073
861 + 164073 = 2a
164934 = 2a
a = 164934 ÷ 2
a= 82467
15^th term = a + ( n - 1 )d
= 82467 + ( 15 - 1) × - 12621
= 82467 + 14 × - 12621
= 82467 - 176694
= - 94227
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