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Question
(a) Write the nth term (Tn) of an Arithmetic Progression (A.P.) consisting of all whole numbers which are divisible by 3 and 7.
(b) How many of these are two-digit numbers? Write them.
(c) Find the sum of first 10 terms of this A.P.
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Solution
Given: (a) Write the nth term (Tn) of an A.P. consisting of all whole numbers divisible by 3 and 7; (b) How many of these are two‑digit numbers? Write them; (c) Find the sum of the first 10 terms of this A.P.
Step-wise calculation:
1. Identify the sequence:
A number divisible by both 3 and 7 is divisible by 1 cm (3, 7) = 21.
The positive whole numbers (natural numbers) divisible by 21 form the A.P.: 21, 42, 63, 84, ... so first term a = 21 and common difference d = 21.
2. (a) nth term Tn:
Use Tn = a + (n – 1)d.
Tn = 21 + (n – 1) × 21 = 21n.
3. (b) Two‑digit terms:
Two‑digit numbers lie between 10 and 99 inclusive.
Solve 10 ≤ 21n ≤ 99
⇒ `n ≥ 10/21 = 1` and `n ≤ 99/21 = 4`
So, n = 1, 2, 3, 4. The two‑digit terms are 21, 42, 63, 84. Count = 4.
4. (c) Sum of first 10 terms S10:
First term a = 21, T10 = 21 × 10 = 210.
Use `S_10 = 10/2 xx (a + T_10)`
= 5 × (21 + 210)
= 5 × 231
= 1155
