What is the first four terms of the A.P. whose first term is $3$ and common difference is $5$ ?

8, 13, - 18, 23 . . . . . . .

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8, 13, 18, 23 . . . . . . .

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8, - 13, 18, 23 . . . . . . .

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- 8, 13, 18, 23 . . . . . . .

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Solution

The correct option is B $8, 13, 18, 23 . . . . . . .$
Let the series be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ ......
$a_{1} = a = 3$
Common difference $= 5$
$a_{2} = a_{1} + d = 3 + 5 = 8$
$a_{3} = a_{2} + d = 8 + 5 = 13$
$a_{4} = a_{3} + d = 13 + 5 = 18$
$a_{5} = a_{4} + d = 18 + 5 = 23$
Therefore, first four terms of this A.P. will be $8, 13, 18, 23 . . . . . .$

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What is the first four terms of the A.P. whose first term is 3 and common difference is 5?

The correct option is B 8,13,18,23.......Let the series be a1,a2,a3,a4,a5 ......a1=a=3Common difference =5a2=a1+d=3+5=8a3=a2+d=8+5=13a4=a3+d=13+5=18a5=a4+d=18+5=23Therefore, first four terms of this A.P. will be 8,13,18,23......

What is the first four terms of the A.P. whose first term is $3$ and common difference is $5$ ?