Question

Q. The first term of an arithmetic sequence is 8 and common difference is 5.

a. Write first three terms of the sequence.

b. Find the algebraic form of the sequence.​

Answer 1

$\blue{\bigstar}$ First three terms of the sequence are $\large\leadsto\boxed{\rm\pink{8 , 13 \: and \: 18}}$

$\blue{\bigstar}$ The algebraic form of the sequence is $\large\leadsto\boxed{\rm\pink{A_n = 5n + 3}}$

• Given:-

First term of an A.P is 8

Common difference is 5

• To Find:-

First three terms of the sequence

Algebraic form of the sequence

• Solution:-

Given that,

First term(a) = 8

Common difference (d) = 5

a.) The first three terms of an A.P will be a, a+d and a+2d .

Hence,

a = 8

a+d = 8 + 5 = 13

a+2d = 8 + 2(5) = 8 + 10 = 18

Therefore, the 1st three terms are 8 , 13 and 18.

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b.) The algebraic form of the sequence will be given by

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{A_n = a + (n-1) d}}}$

➪ $\sf {A_n = 8 + (n-1) 5A}$

➪ $\sf {A_n = 8 + 5n - 5A}$

$★ \bf\red{A_n = 5n + 3}$

Therefore, the algebraic form of the given A.P will be 5n + 3.

Answer 2

• Given:-

First term of an A.P is 8

Common difference is 5

• To Find:-

First three terms of the sequence

Algebraic form of the sequence

• Solution:-

Given that,

First term(a) = 8

Common difference (d) = 5

a.) The first three terms of an A.P will be a, a+d and a+2d .

Hence,

a = 8

a+d = 8 + 5 = 13

a+2d = 8 + 2(5) = 8 + 10 = 18

Therefore, the 1st three terms are 8 , 13 and 18.