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This is one of our many quizzes on arithmetic progression. Click on this to access more such competitive quizzes on arithmetic geometric sequence.

Arithmetic progression Quiz 2

mathematics

 If the sum of 7 terms of an AP is equal to the sum of 15 terms of the AP, the sum of 22 terms of the AP is:

 First find the value of $2a+21d=0$ by the summation formula, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$.

  $\begin{align} & \text{Let the first term be “a” and the common difference is “d”.} \\ & {{S}_{7}}=\dfrac{7}{2}\left[ 2a+6d \right]\text{ and }{{S}_{15}}=\dfrac{15}{2}\left[ 2a+14d \right]\\ & {{S}_{7}}={{S}_{15}}\\ & \Rightarrow \dfrac{7}{2}\left[ 2a+6d \right]=\dfrac{15}{2}\left[ 2a+14d \right]\\ & \Rightarrow 2a+21d=0\\ & {{S}_{22}}=\dfrac{22}{2}\left[ 2a+21d \right]=0\end{align}$  

 If $1 + 5 + 9 + 13 + 17 + ….. + x = 190$,  then what is the value of $x$?

 First find the $a$ and the $d$ from the AP, then by${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, find the value of $x$. 

  $\begin{align} & a = 1, d = 4\\ & \Rightarrow 190=\dfrac{n}{2}\left[ 4n-2 \right]\\ & \Rightarrow 2{{n}^{2}}-n-190=0\\ & \Rightarrow n=10\\ & \Rightarrow x=1+36=37\end{align}$

 If an arithmetic progression has 17 terms, the combined value of the middle three terms of this progression is 180, and the combined value of the last three terms is 390, find out the 5th term.

 First find the ${{9}^{th}}$ term, the common difference from the ${{16}^{th}}$ term then by using the formula for nth term, find the ${{5}^{th}}$ term. 

 $\begin{align} & \text{The middle three terms will be }{{8}^{th}}, {{9}^{th}}\text{ and }{{10}^{th}}\text{ terms.}\\ & \text{The average of these three terms }= {{9}^{th}}\text{ term }= \dfrac{180}{3} = 60.\\ & \text{The average of the last three terms }= {{16}^{th}}\text{ term }= \dfrac{390}{3} = 130.\\ & \text{Now, common difference }= 130 - 60 = \dfrac{70}{7}= 10.\\ & \text{If the }{{9}^{th}}\text{ term is 60, the }{{5}^{th}}\text{ term will be, }60 - 4d\text{ i.e. }60 – 4 \times 10 = 20\end{align}$.

 The ratio of sum of n terms if two AP’s is $\left( 8n+1 \right):\left( 5n+25 \right)$ then the ratio of ${{m}^{th}}$ terms of the AP’s is: 

 Replace n by $\left( 2m1 \right).$, in the formula $\dfrac{2{{a}_{1}}+\left( n-1 \right){{d}_{1}}}{2{{a}_{2}}+(n-1){{d}_{2}}}$.

 $\begin{align} & \text{The ratio,}\dfrac{2{{a}_{1}}+\left( n-1 \right){{d}_{1}}}{2{{a}_{2}}+(n-1){{d}_{2}}}=\dfrac{8n+1}{5n+25}\\ & \text{Replace “n” by }\left( 2m1 \right).\\ & \Rightarrow \dfrac{{{a}_{1}}+\left( m-1 \right){{d}_{1}}}{{{a}_{2}}+\left( m-1 \right){{d}_{2}}}=\dfrac{16m-7}{10m+20}\end{align}$ 

 Find the sum of the first 17 terms of an arithmetic progression where the combined value of the sixth and twelfth term is 14.

 First find the relation between $6^{th}\text{ and }12^{th}$ term then use summation formula, ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ 

 $\begin{align} & \text{The third term }{{t}_{6}}=a+5d\\ & \text{The ninth term }{{t}_{12}}=a+11d\\ & {{t}_{6}}+{{t}_{12}}=2a+16d=14\\ & \text{Sum of first 11 terms of an AP is given by } \\ & {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right],n=17 \\ & {{S}_{17}}=\dfrac{17}{2}\left[ 2a+16d \right]=\dfrac{17}{2}(14)=119 \end{align}$