Question 1 Exercise 4.3

Solutions of Question 1 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Find indicated term and sum of the indicated number of terms in arithmetic sequence: $9,7,5,3, \ldots$; 20th term; 20 terms.

Let $a_1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=9 \\ &d=7-9=-2 \\ &n=20. \end{align} We know that \begin{align}&a_n=a_1+(n-1)d \\ \implies &a_20=9+(20-1)(-2)=-29. \end{align} Assume $S_n$ represents the sum of first $n$ terms of A.P. Then \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \implies S_{20}&=\dfrac{20}{2}[9-29] \\ &=10(-20)=-200 \end{align} Hence 20th term is -29 and sum of first 20 terms is -200.

Find indicated term and sum of the indicated number of terms in case of arithmetic sequence: $3, \dfrac{8}{3}, \dfrac{7}{3}, 2, \ldots$; 11th term; 11 terms.

Let $a_1$ be first term and $d$ be common difference of given A.P. Then \begin{align}&a_1=3 \\ &d=\dfrac{8}{3}-3=-\dfrac{1}{3} \\ &n=11. \end{align} We know that $$a_n=a_1+(n-1) d,$$ This gives $$a_{11}=3+10\left(-\dfrac{1}{3}\right)=-\dfrac{1}{3}$$ Assume $S_n$ represents the sum of first $n$ terms of A.P. Then \begin{align} S_n&=\dfrac{n}{2}[a_1+a_n], \\ \implies S_{11}&=\dfrac{11}{2}\left[3-\dfrac{1}{3}\right] \\ &=\dfrac{11}{2} \cdot \dfrac{9-1}{3} \\&=\dfrac{44}{3} \end{align} Hence 11th term is $-\dfrac{1}{3}$ and sum of first 11 terms is $\dfrac{44}{3}$.

Question 2 >