Advertisements
Advertisements
Question
If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero.
Advertisements
Solution
We know,
\[a_n = a + \left( n - 1 \right)d\]
According to the question,
\[m\left( a_m \right) = n\left( a_n \right)\]
\[ \Rightarrow m\left( a + \left( m - 1 \right)d \right) = n\left( a + \left( n - 1 \right)d \right)\]
\[ \Rightarrow am + \left( m - 1 \right)md = an + \left( n - 1 \right)nd\]
\[ \Rightarrow am + m^2 d - md = an + n^2 d - nd\]
\[ \Rightarrow am - an = n^2 d - nd - m^2 d + md\]
\[ \Rightarrow a\left( m - n \right) = d\left( n^2 - m^2 \right) + d\left( m - n \right)\]
\[ \Rightarrow a\left( m - n \right) = d\left( m + n \right)\left( n - m \right) + d\left( m - n \right)\]
\[ \Rightarrow a\left( m - n \right) = d\left[ \left( m + n \right)\left( n - m \right) + \left( m - n \right) \right]\]
\[ \Rightarrow a\left( m - n \right) = d\left[ - \left( m + n \right)\left( m - n \right) + \left( m - n \right) \right]\]
\[ \Rightarrow a\left( m - n \right) = d\left( m - n \right)\left[ 1 - m - n \right]\]
\[ \Rightarrow a = d\left( 1 - m - n \right) \left( \because m \neq n \right)\]
\[ \Rightarrow a = d\left( 1 - m - n \right) . . . \left( 1 \right)\]
Now,
\[a_{m + n} = \left( a + \left( m + n - 1 \right)d \right)\]
\[ = \left( \left( 1 - m - n \right)d + \left( m + n - 1 \right)d \right) \left( \text{from } \left( 1 \right) \right)\]
\[ = d\left( 1 - m - n + m + n - 1 \right)\]
\[ = 0\]
Hence, the (m + n)th term of the A.P. is zero.
APPEARS IN
RELATED QUESTIONS
Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ...., 185
Find the sum of the following APs:
2, 7, 12, ..., to 10 terms.
Find the sum of the following APs.
0.6, 1.7, 2.8, …….., to 100 terms.
Find the sum of the following arithmetic progressions
`(x - y)^2,(x^2 + y^2), (x + y)^2,.... to n term`
Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ...,
The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.
Determine the nth term of the AP whose 7th term is -1 and 16th term is 17.
Write the next term of the AP `sqrt(2) , sqrt(8) , sqrt(18),.........`
The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.
Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
For an A.P., If t1 = 1 and tn = 149 then find Sn.
Activitry :- Here t1= 1, tn = 149, Sn = ?
Sn = `"n"/2 (square + square)`
= `"n"/2 xx square`
= `square` n, where n = 75
Find the sum of numbers between 1 to 140, divisible by 4
The sum of first n terms of the series a, 3a, 5a, …….. is ______.
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]
If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)
Find the sum of last ten terms of the AP: 8, 10, 12,.., 126.
Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?
Complete the following activity to find the 19th term of an A.P. 7, 13, 19, 25, ........ :
Activity:
Given A.P. : 7, 13, 19, 25, ..........
Here first term a = 7; t19 = ?
tn + a + `(square)`d .........(formula)
∴ t19 = 7 + (19 – 1) `square`
∴ t19 = 7 + `square`
∴ t19 = `square`
Three numbers in A.P. have the sum of 30. What is its middle term?
The sum of A.P. 4, 7, 10, 13, ........ upto 20 terms is ______.
