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प्रश्न
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
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उत्तर
For an A.P.
t4 = 11
`\implies` a + 3d = 11 ...(i)
Also, t8 – 2t4 = 5
`\implies` (a + 7d) – 2 × 11 = 5
`\implies` a + 7d – 22 = 5
`\implies` a + 7d = 27 ...(ii)
Subtracting equation (i) from equation (ii), we get
a + 3d = 11
a + 7d = 27
– – –
– 4d = – 16
d = `(- 16)/(- 4)`
`\implies` d = 4
Substituting d = 4 in equation (i), we get
a + 3d = 11
a + 3 × 4 = 11
`\implies` a + 12 = 11
`\implies` a = –1
∴ Required A.P. = a, a + d, a + 2d, a + 3d, ....
∴ a = –1
∴ a + d = –1 + 4 = 3
∴ a + 2d = –1 + 2(4) = –1 + 8 = 7
∴ a + 3d = –1 + 3(4) = –1 + 12 = 11
Sum of first 50 terms of this A.P.
=`50/2 [2 xx (-1) + 49 xx 4]`
= 25[–2 + 196]
= 25 × 194
= 4850
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