Question

An A.P consists of 50 terms of which the 3rd term is 12 and the last term is 106. Its 29th term is:
(a) 58
(b) 60
(c) 61
(d) 64

Answer 1

Hint: Here, use the formula for the nth term of A.P that is ${{a}_{n}}=a+\left( n-1 \right)d$ to write the last term that is the 50th term and the 3rd term of A.P. From these two equations, find a and d and substitute them in the formula of the 29th term to get the desired answer.

Complete step-by-step answer:
Here, we are given an A.P which has 50 terms of which 3rd term is 12 and the last term is 106. We have to find its 29th term. Let us first understand what an arithmetic progression (A.P) is. Arithmetic Progression is the series of numbers so that the difference of any two successive numbers is a constant value. For example, the series of even numbers: 2, 4, 6, 8…. are in A.P with the first term as 2 and common difference as 2. Also, the nth term of A.P is ${{a}_{n}}=a+\left( n-1 \right)d$ where ‘a’ is the first term, and ‘d’ is the common difference.
Now, we know that the nth term of an A.P is ${{a}_{n}}=a+\left( n-1 \right)d......\left( i \right)$
Also, we are given that the 3rd term of the A.P is 12. So, by substituting n = 3 and ${{a}_{n}}=12$ in the above equation, we get,

& 12=a+\left( 3-1 \right)d \\\ & 12=a+2d \\\ & \Rightarrow a+2d=12....\left( ii \right) \\\ \end{aligned}$$ Also, we are given that there are a total of 50 terms and the last term is 106. This means that the 50th term of an A.P is 106. So, by substituting n = 50 and $${{a}_{n}}=106$$ in equation (i), we get, $$\begin{aligned} & 106=a+\left( 50-1 \right)d \\\ & \Rightarrow 106=a+\left( 49 \right)d \\\ & \Rightarrow a+\left( 49 \right)d=106....\left( iii \right) \\\ \end{aligned}$$ By subtracting equation (ii) from (iii), we get, $$\left( a+49d \right)-\left( a+2d \right)=106-12$$ By simplifying the above equation, we get, $$\begin{aligned} & \Rightarrow 49d-2d=94 \\\ & \Rightarrow 47d=94 \\\ \end{aligned}$$ By dividing 47 on both sides, we get $$d=\dfrac{94}{47}=2$$ So, we get d = 2. By substituting the value of d = 2 in equation (ii), we get, $$\begin{aligned} & a+2\left( 2 \right)=12 \\\ & \Rightarrow a+4=12 \\\ & \Rightarrow a=12-4 \\\ & \Rightarrow a=8 \\\ \end{aligned}$$ So, we get a = 8. Now, we have to find the 29th term. So, by substituting n = 29 in equation (i), we get, $$\begin{aligned} & {{a}_{29}}=a+\left( 29-1 \right)d \\\ & \Rightarrow {{a}_{29}}=a+28d \\\ \end{aligned}$$ By substituting the value of a = 8 and d = 2, we get, $$\begin{aligned} & {{a}_{29}}=8+28\left( 2 \right) \\\ & {{a}_{29}}=64 \\\ \end{aligned}$$ So, we get the 29th term of the A.P as 64. Hence, option (d) is the right answer. Note: In this question, many students make this mistake of writing 3rd term, 29th term, 50th term, etc. as a + 3d, a + 29d, a + 50d respectively but this is wrong. They must consider the formula for the nth term of the A.P that is $${{a}_{n}}=a+\left( n-1 \right)d$$ and write correctly the various terms like 3rd term, 29th term, 50th term as a + 2d, a + 28d and a + 49d respectively where ‘a’ is the first term of A.P and ‘d’ is the common difference of A.P.