Question

Find the \[{{11}^{th}}\]term of an A. P. -27, -22, -17, -12, ....

Answer 1

Hint: In this question, we get the value of a from the first term in the given series. then find the common difference which can be found by finding the difference between two consecutive terms given by the formula \[d={{a}_{2}}-{{a}_{1}}\] . Then find the \[{{11}^{th}}\] term of the series using the formula for \[{{n}^{th}}\] term of an AP which is given by \[{{T}_{n}}=a+\left( n-1 \right)d\]. Now, on further substituting and simplifying we get the result.

Complete step-by-step answer:
ARITHMETIC PROGRESSION (A.P)
A sequence in which the difference of two consecutive terms is constant, is called Arithmetic progression.
Where, a is called the first term of the series and d is called the common difference of the series
Here, the value of the common difference is given by
\[d={{a}_{2}}-{{a}_{1}}\]
As we already know that the \[{{n}^{th}}\] term of an arithmetic progression is given by the formula
\[{{T}_{n}}=a+\left( n-1 \right)d\]
Now, from the given series in the question we have
-27, -22, -17, -12, ....
Now, we have the first term of this series as 18 which gives
\[a=-27\]
Let us now find the common difference of this series
\[\Rightarrow d={{a}_{2}}-{{a}_{1}}\]
Now, on substituting the respective values we get,
\[\Rightarrow d=-22-\left( -27 \right)\]
Now, on further simplification we get,
\[\Rightarrow d=5\]
Now, the \[{{n}^{th}}\]term of the series is given by
\[\Rightarrow {{T}_{n}}=a+\left( n-1 \right)d\]
Here, as we need to find the \[{{11}^{th}}\] term we have
\[n=11\]
Now, on substituting the respective values in the formula we have
\[\Rightarrow {{T}_{11}}=-27+\left( 11-1 \right)5\]
Now, this can be further written in the simplified form as
\[\Rightarrow {{T}_{11}}=-27+50\]
Now, on further simplification we get,
\[\therefore {{T}_{11}}=23\]
Hence, the \[{{11}^{th}}\] term of the given series is 23.

Note: Instead of finding the required term using the \[{{n}^{th}}\]term of arithmetic progression formula we can also find the consecutive terms using the common difference but it would be so lengthy as we need to find till 11 terms.It is important to note that while calculating the common difference we need to subtract the prior term from the later term. It is also to be noted that while simplifying and substituting we should not neglect any of the terms or the sign.