a+1‚3a‚4a+2 are in Ap find a
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First term ( a ) = a + 1
Second term ( a2 ) = 3a
And,
Third term ( a3 ) = 4a + 2
Common difference ( d ) = 3a - ( a + 1 )
=> 3a - a - 1 = 2a - 1
Also,
Common difference ( d ) = 4a + 2 - 3a = a + 2.
We know that,
Common difference of an AP is always equal.
So,
2a - 1 = a + 2
2a - a = 2 + 1
a = 3
First term ( a ) = a + 1
Second term ( a2 ) = 3a
And,
Third term ( a3 ) = 4a + 2
Common difference ( d ) = 3a - ( a + 1 )
=> 3a - a - 1 = 2a - 1
Also,
Common difference ( d ) = 4a + 2 - 3a = a + 2.
We know that,
Common difference of an AP is always equal.
So,
2a - 1 = a + 2
2a - a = 2 + 1
a = 3
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