Question 5 and 6, Exercise 4.3

Solutions of Question 5 and 6 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. $a_{1}=50$, $n=20$, $d=-4$.

Statement is logically incorrect.

Find the sum of arithmetic series with: $a_{1}=50$, $n=20$, $d=-4$.

Solution. Given: $a_{1}=50$, $n=20$, $d=-4$.
Let $S_n$ represents sum of arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \implies S_{20}&=\frac{20}{2}[2(50)+(20-1)(-4)]\\ &=10\times [100-76]\\ &=240. \end{align} Hence $S_{20}=240$.

Find the sum of series. $-3+(-7)+(-11)+\cdots +a_{10}$

Solution. Given series is arithmetic series with $a_1=-3$, $d=-7-(-3)=-4$, $n=10$.
Let $S_n$ represents sum of arithmetic series. Then \begin{align} S_n&=\frac{n}{2}[2a_1+(n-1)d] \\ \implies S_{10}&=\frac{10}{2}[2(-3)+(10-1)(-4)]\\ &=5\times [-6-36]\\ &=-210. \end{align} Hence $S_{10}=-210$.

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