96
(a) 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.
OR
(b) Find the sum of first 20 terms of an A.P. whose $n^{th}$ term is given by $a_n = 5 + 2n$. Can 52 be a term of this A.P.?
OR
(b) Find the sum of first 20 terms of an A.P. whose $n^{th}$ term is given by $a_n = 5 + 2n$. Can 52 be a term of this A.P.?
Show SolutionHide Solution↓
(a) $a + 2d = 16 \dots(i)$ ($\frac{1}{2}$ 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)
OR
(b) $a_n = 5 + 2n \Rightarrow a = 7, d = 2$ (1 mark)
$S_{20} = \frac{20}{2} [14 + 19 \times 2] = 520$ (1 mark)
$52 = 7 + (n - 1) \times 2 \Rightarrow n = \frac{47}{2}$, which is not a natural number. ($\frac{1}{2}$ mark)
Therefore, 52 cannot be a term of this A.P. ($\frac{1}{2}$ 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)
OR
(b) $a_n = 5 + 2n \Rightarrow a = 7, d = 2$ (1 mark)
$S_{20} = \frac{20}{2} [14 + 19 \times 2] = 520$ (1 mark)
$52 = 7 + (n - 1) \times 2 \Rightarrow n = \frac{47}{2}$, which is not a natural number. ($\frac{1}{2}$ mark)
Therefore, 52 cannot be a term of this A.P. ($\frac{1}{2}$ mark)