Question

Question 6
The ${26}^{th},{11}^{th}$ and the last terms of an AP are 0, 3 and $-\frac{1}{5}$, respectively. Find the common difference and the number of terms.

Answer 1

Let the first term, common difference and number of terms of the AP are a, d and n, respectively.
We know that, if last term of an AP is known, then,
l = a + (n - 1)d . . . . . (i)
And nth term of an AP is,
$T_n$ = a + (n - 1)d . . . . . (ii)
Given that, ${26}^{th}$ term of an AP = 0
$\Rightarrow T_{26}=a+(26-1)d=0[fromeq.(i\left)\right]$
$\Rightarrow$ a + 25d = 0 . . . . (iii)
${11}^{th}$ term of an AP = 3
$\Rightarrow T_{11}=a+(11-1)d=3[fromeq.(ii\left)\right]$
$\Rightarrow$ a + 10d = 3 . . . . .(iv)
Last term of the $AP,l=-\frac{1}{5}$
$\Rightarrow$ l = a + (n - 1)d [From eq. (i)]
$\Rightarrow$ $+\frac{1}{5}$ = a + (n - 1)d . . . . . (v)
Now, subtracting eq.(iv) from eq.(iii),
$a+25d=0a+10d=3---\underset{–}{}15d=-3$
$\Rightarrow d=-\frac{1}{5}$
Put the value of d in eq.(iii), we get;
$a+25(-\frac{1}{5})=0\Rightarrow a-5=0\Rightarrow a=5$
Now, put the value of a, d in eq.(v), we get;
$-\frac{1}{5}=5+(n-1)(-\frac{1}{5})$
$\Rightarrow$ -1 = 25 - (n - 1)
$\Rightarrow$ -1 = 25 - n + 1
$\Rightarrow$ n = 25 + 2 = 27
Hence, the common difference and number of term are $-\frac{1}{5}$ and 27, respectively.