Question

(i) Find the sixth term from the end of AP 17, 14, 11,….. – 40.
(ii) Which term of the AP: 5, 15, 25….. will be 130 more than the \[{{31}^{st}}\] term.
(iii) The \[{{8}^{th}}\] term of AP is zero. Prove that \[{{38}^{th}}\] term is triple of \[{{18}^{th}}\] the term.

Answer 1

Hint: To solve the given question we will first find out what an AP is and what the types of terms in this series are. Then we will solve each part of the question separately. To solve part (i), we will first assume that there are n terms such that \[{{a}_{n}}=-40.\] By applying the formula \[{{a}_{n}}=a+\left( n-1 \right)d,\] we will find the value of n. Then, we will find the \[{{\left( n-6+1 \right)}^{th}}\] term of the AP which will be equal to the \[{{6}^{th}}\] term from the end. To solve part (ii), we will first find out the \[{{31}^{st}}\] term of this AP. Then, we will add 130 to it and we will equate it to \[{{r}^{th}}\] term. From here, we will get the value of r. To solve part (iii), we will equate the \[{{8}^{th}}\] term to zero and find the relation between the first term and the common difference. With the help of this relation, we will determine the relation between the \[{{18}^{th}}\] term and the \[{{38}^{th}}\] term.


Complete step-by-step answer:
Before we solve the given question, we must know that AP is a kind of sequence in which the difference between the consecutive terms is constant. This difference is called the common difference. So, let us start solving part (i) of the question.
(i) We can see from the AP that the common difference, in this case, is (– 3). Let us assume that there are n terms in the given AP and \[{{a}_{n}}=-40.\] Now, we know that \[{{r}^{th}}\] term of any AP is given by
\[{{a}_{r}}=a+\left( r-1 \right)d\]
where ‘a’ is the first term and d is the common difference. In our case r = n, a = 17 and d = – 3. Thus,
\[-40=17+\left( n-1 \right)\left( -3 \right)\]
\[\Rightarrow -40-17=-3\left( n-1 \right)\]
\[\Rightarrow -57=-3\left( n-1 \right)\]
\[\Rightarrow \dfrac{-57}{-3}=n-1\]
\[\Rightarrow 19=n-1\]
\[\Rightarrow n=20\]
Thus, there are 20 terms in this AP. The sixth term from the end will be the \[{{15}^{th}}\] term. Thus,
\[\Rightarrow {{a}_{15}}=17+\left( 15-1 \right)\left( -3 \right)\]
\[\Rightarrow {{a}_{15}}=17+\left( 14 \right)\left( -3 \right)\]
\[\Rightarrow {{a}_{15}}=17-42\]
\[\Rightarrow {{a}_{15}}=-25\]
Therefore, the sixth term from the end is – 25.
(ii) By looking at the given AP, we can clearly tell that the d which is the common difference is 10 and the first term is 5. Now, we know that the \[{{r}^{th}}\] term of the AP is given by
\[{{a}_{r}}=a+\left( r-1 \right)d\]
So, the \[{{31}^{st}}\] term will be
\[{{a}_{31}}=5+\left( 31-1 \right)10\]
\[\Rightarrow {{a}_{31}}=5+30\left( 10 \right)\]
\[\Rightarrow {{a}_{31}}=5+300\]
\[\Rightarrow {{a}_{31}}=305\]
Now, we will assume that the \[{{n}^{th}}\] term is 130 more than the \[{{31}^{st}}\] term. Thus,
\[\Rightarrow {{a}_{31}}+130=5+\left( n-1 \right)\left( 10 \right)\]
\[\Rightarrow 305+130=5+\left( n-1 \right)\left( 10 \right)\]
\[\Rightarrow 435=5+\left( n-1 \right)\left( 10 \right)\]
\[\Rightarrow 430=\left( n-1 \right)\left( 10 \right)\]
\[\Rightarrow \left( n-1 \right)=\dfrac{430}{10}\]
\[\Rightarrow \left( n-1 \right)=43\]
\[\Rightarrow n=44\]
Thus, \[{{44}^{th}}\] the term is 130 more than the \[{{31}^{st}}\] term.
(iii) Let us assume that the first term of the given AP is ‘a’ and the common difference is ‘d’. We know that \[{{r}^{th}}\] the term of AP is given by
\[{{a}_{r}}=a+\left( r-1 \right)d\]
In our case, r = 8. Thus, we have,
\[{{a}_{8}}=a+\left( 8-1 \right)d\]
\[\Rightarrow 0=a+7d\]
\[\Rightarrow a=-7d.....\left( i \right)\]
Now, we will find the \[{{18}^{th}}\] term. Thus is given by
\[{{a}_{18}}=a+\left( 18-1 \right)d\]
\[\Rightarrow {{a}_{18}}=a+17d.....\left( ii \right)\]
Now, we will substitute ‘a’ from (i) to (ii). Thus, we will get,
\[\Rightarrow {{a}_{18}}=-7d+17d\]
\[\Rightarrow {{a}_{18}}=10d......\left( iii \right)\]
Similarly, the \[{{38}^{th}}\] term will be
\[{{a}_{38}}=a+\left( 38-1 \right)d\]
\[\Rightarrow {{a}_{38}}=a+37d......\left( iv \right)\]
From (i), we will substitute the value of ‘a’ into (iv). Thus, we will get,
\[\Rightarrow {{a}_{38}}=-7d+37d\]
\[\Rightarrow {{a}_{38}}=30d.....\left( v \right)\]
From (iii) and (v), we can say that,
\[\Rightarrow {{a}_{38}}=3\times 10d\]
\[\Rightarrow {{a}_{38}}=3{{a}_{18}}\]
Hence proved.

Note: We can also solve part (i) by an alternate method as shown. We will reverse the AP given in the question. Thus, the new AP becomes – 40, – 37, …..11, 14, 17. It has a common difference d = 3. The sixth term from the end of the given AP in the question will be the sixth term of the reversed AP. Thus,
\[{{a}_{6}}=a+\left( 6-1 \right)d\]
\[\Rightarrow {{a}_{6}}=-40+\left( 6-1 \right)3\]
\[\Rightarrow {{a}_{6}}=-40+5\left( 3 \right)\]
\[\Rightarrow {{a}_{6}}=-40+15\]
\[\Rightarrow {{a}_{6}}=-25\]