the first term of two aps are equal and the ratios of their common difference is 1:2.if the 7th term of first ap and 21th term of second ap are 23 and 125 respectively.find two aps.
Answers
Step-by-step explanation:
Given The first term of two aps are equal and the ratios of their common difference is 1:2.if the 7th term of first ap and 21th term of second ap are 23 and 125 respectively. find two aps.
According to question we have
- So a7 = 23
- Now a + 6d = 23
- Or a = 23 – 6d------------------1
- Also a21 = 125
- Now a + 20b = 125------------2
- Or 23 – 6d + 20 b = 125 (from 1)
- 20 b – 6d = 102
- 10 b – 3d = 51
- Now according to question d/b = 1/2
- Now d = 1x and b = 2x (for some positive value x)
- Now 10 b – 3d = 51
- 10(2x) – 3 (x) = 51
- 17 x = 51
- Or x = 3
- Therefore d = 3, b = 6
- Put d = 3 in equation 1 we get
- So a + 6(3) = 23
- Or a = 23 – 18
- Or a = 5
Therefore first A.P is 5,8,11-------and
Second A.P is 5,11,17-------
Reference link will be
https://brainly.in/question/15416700
First arithmetic progressions is:
5, 8, 11, 14, 17, 20, 23, 25, ....
Second arithmetic progressions is:
5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, 125, 131, .....
Step-by-step explanation:
The arithmetic progression is given as:
an = a1 + (n - 1)d
Where,
a1 = First term
d = Common difference
n = nth term
From question, 7th term is:
a7 = 23
a + 6d = 23
⇒ a = 23 - 6d → (equation 1)
a21 = 125
a + 20d = 125
⇒ a = 125 - 20b → (equation 2)
On equating equation (1) and (2), we get,
23 - 6d = 125 - 20b
23 - 6d + 20b = 125
20b - 6d = 125 - 23
20b - 6d = 102 → (equation 3)
On dividing equation 3 by 2, we get,
10b - 3d = 51 → (equation 4)
From question, d/b = 1/2 ⇒ d = 1x and b = 2x
On substituting d and b in equation (4), we get,
10(2x) - 3(x) = 51
20x - 3x = 51
17x = 51
∴ x = 3
⇒ d = 3 and b = 6
⇒ a = 5