Question

if n-2 , 4n-1 and 5n+2 are in A.P.Find n
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Answer 1

Question

if n-2 , 4n-1 and 5n+2 are in A.P.Find n

Given

$\sf{ \to \: a_{1} = n - 2} \\ \sf{ \to \: a_{2} = 4n - 1} \\ \sf{ \to \: a_{3} = 5n + 2}$

Solution

$\sf{a_{2} - a_{1} = a_{3} - a_{2} = d}$

where

$\sf{a_{1} = first \: term } \\ \sf{a_{2} = second \: term} \\ \sf{a_{3} = third \: term} \\ \sf{ d = common \: difference}$

substitute the values

4n-1 -n +2 = 5n+2 - 4n+1

3n +1 = n +3

2n = 2

n= 2/2 = 1

$\sf{ \therefore \: the \: vaue \: of \: n = 1}$

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More to know !!

Sum of n terms in AP

S = n/2[2a + (n − 1) × d]

General Form of AP

a, a + d, a + 2d, a + 3d, . . .

The nth term of AP

an = a + (n – 1) × d

Sum of all terms in a finite AP with the last term as 'l'

n/2(a + l)

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