Question

The ${11^{th}}$ term and the ${21^{st}}$ term of an AP are 16 and 29 respectively, then find the ${34^{th}}$ term.

Answer 1

Hint: Here, we will solve the given problem using the ${n^{th}}$ term formulae of an AP i.e., ${T_n} = a + (n - 1)d$.

Complete step-by-step answer:
Now the ${11^{th}}$ term and ${21^{st}}$ term is given to us.
That is, ${t_{11}} = 16$ and ${t_{21}} = 29$.
Now any ${n^{th}}$ term of an AP can be written as ${T_n} = a + (n - 1)d \to (1)$
Here ‘a’ is the first term and ‘d’ is the common difference.
Using eq 1 we can write
${t_{11}} = a + (11 - 1)d \Rightarrow 16 = a + (11 - 1)d$
${t_{21}} = a + (21 - 1)d \Rightarrow 29 = a + (21 - 1)d$
So the two equations that we are getting are,
$\Rightarrow$ $16 = a + 10d$ and $29 = a + 20d$
Subtracting both, we get,
$\Rightarrow$ $10d = 13 \Rightarrow d = 1.3$
Putting the value of d in $16 = a + 10d$
We get $16 = a + 10 \times (1.3) \Rightarrow a = 3$
We have to find the ${34^{th}}$ term therefore using eq 1
${t_{34}} = a + (34 - 1)d$
Putting the values of a and d, we get
$\Rightarrow$ ${t_{34}} = 3 + (34 - 1) \times 1.3 = 45.9$

Note: While solving such AP problems, always remember the concept of writing any ${n^{th}}$ term of an AP, this helps simplify and get you on the right track to obtain the answer.