
The nth term of an A.P is $3 n-1$.Choose from the following the sum of its first five terms
A. 14
B. 35
C. 80
D. 40
Answer
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Hint:In this question, we have given the value of nth term of an A.P. Then by using the formula of A.P. series we find the first two terms of an AP series and using this we determine the value of common difference. Then by using the formula of the sum of the series and putting the values, we are able to find the sum of the first five terms of an AP series.
Formula used:
$a_{n}=a+(n-1) \times d$
$d=a_2-a_1$
$S_n=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Complete Step-by-step Solution:
Given that
nth term of AP is $3 n-1$
The formula for the nth term of an AP is
$a_{n}=a+(n-1) \times d$
Where,
a=first term
$a_{n}$=nth term
n=number of terms
$\mathrm{d}=$Common difference
The difference between the second and first terms, is known as the common difference "$d$," and a is the first term of the given AP series.
for $1 \mathrm{st}$ term, $\mathrm{n}=1$
$a_1=3 \times 1-1$ ( putting the value of $n$ )
$a_1=3-1$
$a_1=2$
Therefore 1 st term $=2$
for 2 nd term, $n=2$
$a_2=3 \times 2-1$ ( putting the value of $n$ )
$a_2=6-1$
$a_2=5$
Therefore the 2nd term is 5
Now, $d=a_2-a_1$
$\Rightarrow~d=5-2$
$d=3$
Now, $n=5$ ( since we need to find the first 5 terms)
Now the formula for sum -
$S_n=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Putting the values in the formula - -
$S_n=\dfrac{5}{2}\left[ 2(2)+\left( 5-1 \right)3 \right]$
Then $S_n=\dfrac{5}{2}\left[ 4+\left( 4 \right)3 \right]$
$AP=\dfrac{5}{2}\left[ 4+12 \right]$
$AP=\dfrac{5}{2}\times 16$
$AP=5\times 8$
$AP=40$
So the correct answer is option D.
Note: Students can also solve this question by determining all the values of first term to fifth term. Adding all the values till the fifth term, they can get the sum till 5th term of the given AP. This way can be used when they don’t remember the formulas.
Formula used:
$a_{n}=a+(n-1) \times d$
$d=a_2-a_1$
$S_n=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Complete Step-by-step Solution:
Given that
nth term of AP is $3 n-1$
The formula for the nth term of an AP is
$a_{n}=a+(n-1) \times d$
Where,
a=first term
$a_{n}$=nth term
n=number of terms
$\mathrm{d}=$Common difference
The difference between the second and first terms, is known as the common difference "$d$," and a is the first term of the given AP series.
for $1 \mathrm{st}$ term, $\mathrm{n}=1$
$a_1=3 \times 1-1$ ( putting the value of $n$ )
$a_1=3-1$
$a_1=2$
Therefore 1 st term $=2$
for 2 nd term, $n=2$
$a_2=3 \times 2-1$ ( putting the value of $n$ )
$a_2=6-1$
$a_2=5$
Therefore the 2nd term is 5
Now, $d=a_2-a_1$
$\Rightarrow~d=5-2$
$d=3$
Now, $n=5$ ( since we need to find the first 5 terms)
Now the formula for sum -
$S_n=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Putting the values in the formula - -
$S_n=\dfrac{5}{2}\left[ 2(2)+\left( 5-1 \right)3 \right]$
Then $S_n=\dfrac{5}{2}\left[ 4+\left( 4 \right)3 \right]$
$AP=\dfrac{5}{2}\left[ 4+12 \right]$
$AP=\dfrac{5}{2}\times 16$
$AP=5\times 8$
$AP=40$
So the correct answer is option D.
Note: Students can also solve this question by determining all the values of first term to fifth term. Adding all the values till the fifth term, they can get the sum till 5th term of the given AP. This way can be used when they don’t remember the formulas.
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