Advertisements
Advertisements
प्रश्न
Two A.P.’s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. Show that the difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
उत्तर
Let the two A.P.s be
A.P.1 = a1, a1 + d, a1 + 2d, ...
A.P.2 = a2, a2 + d, a2 + 2d, ...
In A.P.1 we have a1 = 2
In A.P.2 we have a2 = 7
t10 in A.P.1 = a1 + 9d = 2 + 9d ...(1)
t10 in A.P.2 = a2 + 9d = 7 + 9d ...(2)
The difference between their 10th terms
= (1) – (2) = 2 + 9d – 7 – 9d
= – 5 ...(I)
t21 m A.P.1 = a1 + 20d = 2 + 20d ...(3)
t21 in A.P.2 = a2 + 20d = 7 + 20d ...(4)
The difference between their 21st terms is
(3) – (4)
= 2 + 20d – 7 – 20d
= – 5 ...(II)
I = II
Hence it is Proved.
APPEARS IN
संबंधित प्रश्न
An arithmetic progression 5, 12, 19, …. has 50 terms. Find its last term. Hence find the sum of its last 15 terms.
The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.
Find the arithmetic progression whose third term is 16 and the seventh term exceeds its fifth term by 12.
Find the 8th term from the end of the A.P. 7, 10, 13, ..., 184
In an A.P., if the 12th term is −13 and the sum of its first four terms is 24, find the sum of its first ten terms ?
Find the common difference of the A.P. and write the next two terms \[0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, . . . \]
Check whether the following sequence is in A.P.
1, –1, 1, –1, 1, –1, …
Find the middle term(s) of an A.P. 9, 15, 21, 27, …, 183.
The list of numbers – 10, – 6, – 2, 2,... is ______.
First term and common difference of an A.P. are 1 and 2 respectively; find S10.
Given: a = 1, d = 2
find S10 = ?
Sn = `n/2 [2a + (n - 1) square]`
S10 = `10/2 [2 xx 1 + (10 - 1)square]`
= `10/2[2 + 9 xx square]`
= `10/2 [2 + square]`
= `10/2 xx square = square`