Question

How do you find the $37th$ term of the sequence \left\\{ 4,7,10,.... \right\\}?

Answer 1

From the given question we have to find the $37$th term of the sequence \left\\{ 4,7,10,.... \right\\}. This question is based on progression, the given sequence is in arithmetic progression so we know the formula for the nth term in an arithmetic progression ${{T}_{n}}=a+\left( n-1 \right)d$ where a is the first term of the sequence, d is the common difference of the sequence and n is the required term. By this formula we will get the required answer.

Complete step by step solution:
From the question, given sequence is \left\\{ 4,7,10,.... \right\\}
In the sequence first term is $4$so by this we can say that
$\Rightarrow a=4$
Now we have to find the common difference in the sequence, the difference between the second term and first term is $3$and the difference between the third term and second term is also $3$.
Therefore, the common difference is $3$, so by this we can say that
$\Rightarrow d=3$
Now we have to find the $37th$ term of the sequence \left\\{ 4,7,10,.... \right\\}
Here n is $37$, so we can say that
$\Rightarrow n=37$
we know the formula for nth term in the arithmetic progression
$\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d$
by substituting the above values
$\Rightarrow {{T}_{37}}=4+\left( 37-1 \right)3$
$\Rightarrow {{T}_{37}}=4+\left( 36 \right)3$
$\Rightarrow {{T}_{37}}=4+\left( 36 \right)3$
$\Rightarrow {{T}_{37}}=4+108$
$\Rightarrow {{T}_{37}}=112$
Therefore, the $37$th term of the given sequence \left\\{ 4,7,10,.... \right\\} is $112$

Note: Students should recall the concepts of arithmetic progression while solving the question. Students should know the difference between arithmetic progression, geometric progression and harmonic progression. Here harmonic progression is formed by taking the reciprocals of an arithmetic progression. Students should know the below formulas of arithmetic progression they are nth term of the arithmetic progression and sum of n terms of the arithmetic progression
$\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d$
$\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$