Question

The algebraic form of an arithmetic sequence is 4 n+1 .

a) What is its common difference ?

b) What is its first term ?

c) What is the remainder when each term of this sequence is divided by 4 ?​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{First\:term=5}}}$

$\green{\tt{\therefore{Common\:difference=4}}}$

$\green{\tt{\therefore{Remainder=1}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green {\underline \bold{Given :}} \\ \tt: \implies a_{n} = 4n + 1 \\ \ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies First \: term = ? \\ \\ \tt: \implies Common \: difference = ? \\ \\ \tt: \implies Remainder \: when \: each \: term \: is \: divided \: by \: 4 = ?$

• According to given question :

$\tt \circ \: a_{n} = 4n + 1 \\ \\ \tt\because \:n = 1,2,3,.... \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies a_{1} = 4 \times 1 + 1 \\ \\ \tt: \implies a_{1} =4 + 1 \\ \\ \green{\tt: \implies a_{1} =5} \\ \\ \bold{For \: n = 2} \\ \tt: \implies a_{2} =4 \times 2 + 1 \\ \\ \tt: \implies a_{2} =8 + 1 \\ \\ \green{\tt: \implies a_{2} = 9} \\ \\ \bold{For \: n = 3} \\ \tt: \implies a_{3} = 4 \times 3 + 1 \\ \\ \tt: \implies a_{3} =12 + 1$

$\green{\tt: \implies a_{3} =13} \\ \\ \bold{For \: common \: difference} \\ \tt: \implies d= a_{2} - a_{1} \\ \\ \tt: \implies d = 9 - 5 \\ \\ \green{\tt: \implies d = 4} \\ \\ \bold{For \: Remainder : } \\ \tt: \implies 5 \div 4 = 4\times 4+1 \\ \\ \tt: \implies Remainder = 1 \\ \\ \tt: \implies 9 \div 4 = 4\times 2+1 \\ \\ \tt: \implies Remainder = 1 \\ \\ \tt: \implies 13 \div 4 = 4\times 3+1 \\ \\ \green{\tt: \implies Remainder = 1} \\ \\ \green{ \tt\therefore For \: all \: term \: of \: this \:A.P\: when\:divide\:by} \\ \: \: \green{\tt 4 \: then \: remainder \: gives \: 1}$

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

$</p><p>\tt{\purple{a_{n} = 4n \: + \: 1 }}$

$</p><p>\tt{\red{\therefore{a_{1} = \: 5 }}}.............(1)$

$</p><p>\tt{\orange{\therefore{a_{2} = 9 }}} ...............(2)$

Hence

$</p><p></p><p>\tt{\red{\bullet{\boxed{\boxed{a = 5 }}}}}$

$</p><p></p><p>\tt{\blue{\bullet{\boxed{\boxed{d = 4 }}}}}$

Each form of the number is 4k + 1.

Divided by 4, we get remainder as 1.