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If 4m+8,2m²+3m+6 & 3m² + 4m …

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If 4m+8,2m²+3m+6 & 3m² + 4m +4 are 3 concecutive terms of an AP. Find m.

Posted by Karuna Karuna 8 years ago

CBSE > Class 10 > Mathematics

1 answers

Rajeev Malhotra 8 years ago

First term = 4m+8

Second term = {tex}2m^2+3m+6{/tex}

Third term = {tex}3m^2+4m+4 {/tex}

Common Difference d = second term - first term

d = {tex}(2m^2+3m+6) - (4m+8){/tex}

d = {tex}2m^2-m-2{/tex}

Common Difference = third term - second term

d = {tex}(3m^2+4m+4)-(2m^2+3m+6){/tex}

d = {tex}m^2+m-2{/tex}

So,

{tex}2m^2-m-2 = m^2+m-2{/tex}

{tex}m^2-2m=0{/tex}

m(m-2)= 0

This gives m= 0 and m-2=0

m=0 or m=2

0Thank You

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CBSE > Class 10 > Mathematics