Question 1 and 2, Exercise 4.3

Solutions of Question 1 and 2 of Exercise 4.3 of Unit 04: Sequence and Series. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Find the sum of series: $4+7+10+13+16+19+22+25$

Solution. Given: $4+7+10+13+16+19+22+25$.

As the given series is arithmetic series with $a_1=4$, $d=7-4=3$, $n=8$, so \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d]\\ \implies S_8&=\frac{8}{2}[2(4)+(8-1)(3)]\\ &=4[8+7\times 3] = 116 \end{align} Hence sum of given series is 116.

Find the sum of series. $a_{1}=2$, $a_{n}=200$, $n=100$

Statement is logically incorrenct.

Find the sum of arithematic series with: $a_{1}=2$, $a_{n}=200$, $n=100$

Solution.

Given $a_{1}=2$, $a_{n}=200$, $n=100$.
Let $S_n$ represents sum of arithematic series. Then \begin{align} S_n&=\frac{n}{2}[a_1+a_n] \\ \implies S_{100}&=\frac{100}{2}[2+200]\\ &=10100. \end{align} Hence $S_{100}=10100$.

Question 3 & 4 >