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Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.
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$a + 2d = 16 \dots(i)$ ($\frac{1}{2}$ mark)
$a + 6d = 12 + a + 4d \dots(ii)$ ($\frac{1}{2}$ mark)
Solving (i) and (ii) to get $d = 6, a = 4$ ($\frac{1}{2}$ mark)
$\therefore \text{A.P. is } 4, 10, 16, \dots$ ($\frac{1}{2}$ mark)
$S_{29} = \frac{29}{2} [8 + 28 \times 6] = 2552$ ($\frac{1}{2} + \frac{1}{2}$ marks)
$a + 6d = 12 + a + 4d \dots(ii)$ ($\frac{1}{2}$ mark)
Solving (i) and (ii) to get $d = 6, a = 4$ ($\frac{1}{2}$ mark)
$\therefore \text{A.P. is } 4, 10, 16, \dots$ ($\frac{1}{2}$ mark)
$S_{29} = \frac{29}{2} [8 + 28 \times 6] = 2552$ ($\frac{1}{2} + \frac{1}{2}$ marks)