====== 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. =====Question 5===== Find the sum of series. $a_{1}=50$, $n=20$, $d=-4$. FIXME 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 $S_{20}=240$. GOOD =====Question 6===== 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 $S_{10}=-210$. GOOD ====Go to ==== [[math-11-nbf:sol:unit04:ex4-3-p2|< Question 3 & 4]] [[math-11-nbf:sol:unit04:ex4-3-p4|Question 7 & 8 >]]