Question

The 11^th term and the 21^st term of an A.P. are 16 and 29 respectively then find
(i) The 1^st term and the common difference
(ii) The 34^th term
(iii) 'n' such that t_n = 55.

Answer 1

We know that the n^th term of an A.P. is t_n = a + (n – 1)d.
Here,
t₁₁ = 16
Thus, we have:
t₁₁ = a + (11 – 1)d = 16
$\Rightarrow$a + 10d = 16 …(1)

We know: t₂₁ = 29
Thus, we have:
t₂₁ = a + (21 – 1)d = 29
$\Rightarrow$a + 20d = 29 …(2)

(i)
On subtracting (2) from (1), we get:
a + 10d – a – 20d = 16 – 29
$\Rightarrow$–10d = – 13
$\Rightarrow$d = $\frac{13}{10}$ = 1.3

On putting d = 1.3 in (1), we get:
a + 10d = 16
$\Rightarrow$a + 10(1.3) = 16
$\Rightarrow$a + 13 = 16
$\Rightarrow$a = 16 – 13 = 3

Thus, the first term of the A.P. is 3 and the common difference is 1.3.

(ii)
Now, we have:
a = 3 and d = 1.3
34^th term of the given A.P.:
t_n = a + (n – 1)d
$\Rightarrow$t₃₄ = 3 + (34 – 1)(1.3)
$\Rightarrow$t₃₄ = 3 + 33 × 1.3
$\Rightarrow$t₃₄ = 3 + 42.9 = 45.9

(iii)
We have: t_n = 55
$\Rightarrow$a + (n – 1)d = 55
We know: a = 3 and d = 1.3
Thus, we have:
3 + (n – 1)(1.3) = 55
$\Rightarrow$3 + 1.3n – 1.3 = 55
$\Rightarrow$1.7 + 1.3n = 55
$\Rightarrow$1.3n = 55 – 1.7 = 53.3
$\Rightarrow$n = $\frac{53.3}{1.3}=41$