Question

Match the following columns:

Column I	Column II
(a) In an AP, it is given that T5 = −9 and T9 = 7. Then, T14 = ?	(p) 8
(b) In an AP, Sn = (n2 + 3n). Then, T16 = ?	(q) 34
(c) In an AP, T10 = 44 and T15 = 64. Then, its first term is	(r) 15
(d) How many 2-digit numbers are divisible by 6?	(s) 27

Answer 1

(a) - (s), (b) - (q), (c) - (p ), (d) - ( r)


(a) T₅ = a + 4d = −9 ...(i)
T₉ = a + 8d = 7 ...(ii)
Subtracting (i) from (ii), we get:
4d = 16
∴ d = 4
On putting the value of d in (i), we get a = −25.
∴ T₁₄ = a + 13d = −25 + 13 ⨯ 4 = −25 + 52 = 27


(b) S_n = (n² + 3n) ...(i)
Replacing n by (n − 1) in (i), we get;
S_n-1 = (n − 1)² + 3(n − 1)
= (n² − 2n + 1) + 3n − 3
= n² + n − 2
∴​ T_n = (S_n − S_n-1)
​ = (n² + 3n) − (n² + n − 2) = 2n + 2
∴ n^th term, T_n = (2n + 2) ...(ii)
Putting n = 16 in (ii), we get:
T₁₆= (2 ⨯ 16) + 2 = 34
Hence, the 16^th term is 34.

(c) T₁₀ = a + 9d = 44 ...(i)
T₁₅ = a + 14d = 64 ...(ii)
Subtracting (i) from (ii), we get:
5d = 20
∴ d = 4
On putting the value of d in (i), we get a = 8.

(d) All the 2-digit numbers divisible by 6 are 12, 18, 24, 30, ..., 96.
This is an AP whose first term is 12 and the common difference is 6.
Let 96 be the n^th term of the AP.
Now, T_n = 96
⇒ a + (n − 1)d = 96
⇒ 12 + (n -1) 6 = 96
⇒ (n − 1) = 14
⇒ n = 15