Question

Find the ${{6}^{th}}$ term from the end of the A.P. 17, 14, 11,………….., -40?

Answer 1

Hint:Try to figure out that the A.P. that you get when you write the above mentioned A.P. in reverse order. Once you figure it out, find the sixth term.

Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{r}}$, and sum till n terms is denoted by ${{S}_{r}}$ .
${{T}_{r}}=a+\left( r-1 \right)d$
${{S}_{r}}=\dfrac{r}{2}\left( 2a+\left( r-1 \right)d \right)$
Now moving to the sequence that you get when you reverse the A.P. given in the question. We would get an A.P. again, which will have a common difference of 3 and the first term of the sequence will be -40. Therefore, the sequence is: -40, -37, -34……………..
From above series $a= -40$ , $d=-37-(-40)=3$
Using above formula the sixth term of the new A.P. is:
${{T}_{6}}=a+\left( 6-1 \right)d$
$\Rightarrow {{T}_{6}}=-40+\left( 6-1 \right)\times 3=-25$
Therefore, the ${{6}^{th}}$ term from the end of the A.P. 17, 14, 11,………….., -40 is -25.

Note: In a question related to an arithmetic progression, the most important thing is to find the common difference and the first term. You also need to learn the basic properties of all the standard sequences like the arithmetic progression and geometric progressions.For these types of questions try to make reverse order of given A.P and find the given term.