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Question: The sum of first m terms of A.P. is \(4{m^2} - m\) . If the nth term is 107, find the value of n. Al...

The sum of first m terms of A.P. is 4m2−m4{m^2} - m . If the nth term is 107, find the value of n. Also, find the 21st term of this A.P.

Explanation

Solution

Hint- In this question we have sum of first m terms of an A.P. so we will determine first term and common difference by putting the value of m as 1 or 2, Now to proceed further we will use the formula of nth term of A.P. as an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d by using it we will get the value of n and again by putting the value of n as 21 in this formula we will get the required result.

Complete step by step answer:
Given that the sum of first m terms of an A.P. is 4m2−m4{m^2} - m
Now to determine the first term put the value of m as 1, because the sum of the first term is first term.
S1=4m2−m S1=4×(1)2−1 S1=3  {S_1} = 4{m^2} - m \\\ {S_1} = 4 \times {\left( 1 \right)^2} - 1 \\\ {S_1} = 3 \\\
Therefore, the value of first term is 3
a=S1=3a = {S_1} = 3
Also, to determine common difference, the sum of first two terms is given by
Putting the value of m as 2 in the given sum equation
S2=4m2−m S2=4×(2)2−2 S2=16−2 S2=14  {S_2} = 4{m^2} - m \\\ {S_2} = 4 \times {\left( 2 \right)^2} - 2 \\\ {S_2} = 16 - 2 \\\ {S_2} = 14 \\\
As we know that
S2=a1+a2=a+a+d{S_2} = {a_1} + {a_2} = a + a + d
Now substituting the value of S2=14,a=3{S_2} = 14,a = 3 in the above equation, we get
S2=a+a+d ⇒14=3+3+d ⇒d=14−6 ⇒d=8  {S_2} = a + a + d \\\ \Rightarrow 14 = 3 + 3 + d \\\ \Rightarrow d = 14 - 6 \\\ \Rightarrow d = 8 \\\
It is given that an=107{a_n} = 107
As we know that the nth term is given as an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d
Substituting the value of an,a and d{a_n},a{\text{ and }}d in the above nth term equation
an=a+(n−1)d ⇒107=3+(n−1)×8 ⇒107−3=8n−8 ⇒104+8=8n ⇒8n=112 ⇒n=1128 ⇒n=14  {a_n} = a + \left( {n - 1} \right)d \\\ \Rightarrow 107 = 3 + \left( {n - 1} \right) \times 8 \\\ \Rightarrow 107 - 3 = 8n - 8 \\\ \Rightarrow 104 + 8 = 8n \\\ \Rightarrow 8n = 112 \\\ \Rightarrow n = \dfrac{{112}}{8} \\\ \Rightarrow n = 14 \\\
Now to find the value of a21{a_{21}}
Here the value of n is 21
Using the nth term formula and substituting the value of a, n and d we get
an=a+(n−1)d a21=3+(21−1)×8 a21=3+(20)×8 a21=3+160 a21=163  {a_n} = a + \left( {n - 1} \right)d \\\ {a_{21}} = 3 + \left( {21 - 1} \right) \times 8 \\\ {a_{21}} = 3 + \left( {20} \right) \times 8 \\\ {a_{21}} = 3 + 160 \\\ {a_{21}} = 163 \\\
Hence, the value of 21 terms is 163163 and the value of n is 14.

Note- In order to solve these types of questions first of all remember the formula of the nth term of A.P. Also remember the formula of sum of n terms of an A.P. In this question the sum of m terms of an A.P is given and from the sum we calculated the value of a and d using the nth term formula. . Similarly learn about geometric progression and harmonic progression. This will help a lot to solve problems related to A.P, G.P and H.P.