
If the tenth term of an A.P. is 21, and the sum of first ten terms is 120, find its nth term.
(a) 7n + 1
(b) 5n + 1
(c) 2n + 1
(d) None of these
Answer
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Hint: We know that the nth term of an arithmetic progression is $n\text{th term}=a+\left( n-1 \right)d$ and the sum of first n terms of an arithmetic progression is given as $\text{Sum of }n\text{ terms}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ . We must use these two equations and solve them simultaneously to get the values of the first term and the common difference. We can then substitute these two values in the general equation for the nth term to get the required answer.
Complete step by step solution:
Let us assume that the first term of this arithmetic progression is a and the common difference is d.
Then, we know that the nth term of an arithmetic progression is given as,
$n\text{th term}=a+\left( n-1 \right)d$
where n is the number of terms.
So, for the tenth term, n = 10.
Hence, we have
$21=a+\left( 10-1 \right)d$
Thus, we get
$a+9d=21$
We also know that the sum of n terms of an arithmetic progression is given by
Sum of n terms = $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Thus, for sum of 10 terms, we have
$120=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right]$
On simplifying, we get
$120=5\left[ 2a+9d \right]$
We can rewrite this as
$\dfrac{120}{5}=2a+9d$
Or, $2a+9d=24$
We can further split 2a as (a + a).
$a+\left( a+9d \right)=24$
Putting the value of (a + 9d) from equation (i), we get
$a+21=24$
Hence, we get the value $a=3$ .
Putting the value of a in equation (i), we get
$3+9d=21$
Simplifying the above equation, we get
$9d=18$
Or, $d=2$.
So, now we have the value of the first term and the common difference. Substituting these values into the general form for nth term, we get
$n\text{th term}=a+\left( n-1 \right)d$
$\Rightarrow n\text{th term}=3+\left( n-1 \right)2$
$\Rightarrow n\text{th term}=2n+1$
Thus, the nth term of this arithmetic progression is 2n + 1.
So, the correct answer is “Option C”.
Note: We can also use the alternate formula for the sum of first n terms as $\text{Sum of }n\text{ terms}=\dfrac{n}{2}\left[ a+l \right]$ , where l is the last term, i.e., the tenth term in this case. Hence, we can directly write $120=5\left[ a+21 \right]$ and calculate a very easily. Using this, we can calculate d, and hence the answer.
Complete step by step solution:
Let us assume that the first term of this arithmetic progression is a and the common difference is d.
Then, we know that the nth term of an arithmetic progression is given as,
$n\text{th term}=a+\left( n-1 \right)d$
where n is the number of terms.
So, for the tenth term, n = 10.
Hence, we have
$21=a+\left( 10-1 \right)d$
Thus, we get
$a+9d=21$
We also know that the sum of n terms of an arithmetic progression is given by
Sum of n terms = $\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Thus, for sum of 10 terms, we have
$120=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right]$
On simplifying, we get
$120=5\left[ 2a+9d \right]$
We can rewrite this as
$\dfrac{120}{5}=2a+9d$
Or, $2a+9d=24$
We can further split 2a as (a + a).
$a+\left( a+9d \right)=24$
Putting the value of (a + 9d) from equation (i), we get
$a+21=24$
Hence, we get the value $a=3$ .
Putting the value of a in equation (i), we get
$3+9d=21$
Simplifying the above equation, we get
$9d=18$
Or, $d=2$.
So, now we have the value of the first term and the common difference. Substituting these values into the general form for nth term, we get
$n\text{th term}=a+\left( n-1 \right)d$
$\Rightarrow n\text{th term}=3+\left( n-1 \right)2$
$\Rightarrow n\text{th term}=2n+1$
Thus, the nth term of this arithmetic progression is 2n + 1.
So, the correct answer is “Option C”.
Note: We can also use the alternate formula for the sum of first n terms as $\text{Sum of }n\text{ terms}=\dfrac{n}{2}\left[ a+l \right]$ , where l is the last term, i.e., the tenth term in this case. Hence, we can directly write $120=5\left[ a+21 \right]$ and calculate a very easily. Using this, we can calculate d, and hence the answer.
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