Question

Question 14
If the nth terms of the AP’s 9 , 7, 5, …. and 24, 21, 18 … are the same, then find the value of n, Also that term.

Answer 1

Let the first term, common difference and number of terms of the AP 9, 7, 5, …. are $a_1,d_{1}and{n}_1$, respectively.
i.e., first term $\left(a_1\right)$ = 9 and common difference $\left(d_1\right)$ = 7 – 9 = - 2
$\therefore$ Its nth term, $T_{n_1}^{\prime }=a_1+(n_1-1)d_1$
$\Rightarrow T_{n_1}^{\prime }=9+(n_1-1)(-2)$
$\Rightarrow T_{n_1}^{\prime }=9-2n_1+2$
$\Rightarrow T_{n_1}^{\prime }=11-2n_1$
[$\because$ nth term of an $AP$, $T_n$ = a + (n - 1)d] . . . . . (i)
Let the first term, common difference and the number of terms of the AP 24, 21, 18, ….. are $a_2,d_{2}and{n}_2$ respectively.
i.e., first term, $\left(a_2\right)$ = 24 and common difference $\left(d_2\right)$ = 21 – 24 = - 3
$\therefore Itsnthterm,T_{n_2}^{\prime \prime }=a_2+(n_2-1)d_2$
$\Rightarrow T_{n_2}^{\prime \prime }=24+(n_2-1)(-3)$
$\Rightarrow T_{n_2}^{\prime \prime }=24-3n_2+3$
$\Rightarrow T_{n_2}^{\prime \prime }=27-3n_2\cdots \left(ii\right)$
Now, by given condition,
nth terms of the both APs are same, i.e., $T_n^{\prime }=T_n^{\prime \prime }$
$11-2n=27-3n$ [ from eqs. (i) and (ii) ]
$\Rightarrow$ n = 16
$\therefore$ nth term of first $AP,T^{\prime }{n}_1=11-2n_1=11-2\left(16\right)$
= 11 – 32 = - 21
nth term of second $AP,T^{\prime \prime }{n}_2=27-3n_2=27-3\left(16\right)$
= 27 – 48 = - 21
Hence, the value of n is 16 and that term i.e., nth term is – 21.