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The sum of 10 terms of the arithmetic series is 390. If the third term of the series is 19, find the first term
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- 3
- 5
- 7
- 8
- 3
Correct Option: A
Here , S10 = 390 and first term ( a ) = ?
Let the first term of A.P. be ‘a’ and the common difference be ‘d’.
| ∴ Sn = | [2a + (n - 1)d] | |
| 2 |
| ⇒ 390 = | [2a + (10 - 1)d] | |
| 2 |
⇒ 390 = 5 (2a + 9d)
| ⇒ 2a + 9d = | = 78 .....(i) | |
| 5 |
Again, third term = 19
[ ∴ Tn = a + (n – 1)d]
⇒ a + 2d = 19 .... (ii)
By equation (i) – 2 × (ii),
2a + 9d – 2a – 4d = 78 – 38
⇒ 5d = 40
| ⇒ d = | = 8 | |
| 5 |
From equation (ii),
a + 2 × 8 = 19
⇒ a = 19 – 16 = 3
