Question

Q26. The value of 'm' for which
( 8m +3), (6 m - 2) and ( 2 m +
7) are in A.P. is ......
15/2
5
7/2
7
​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• (8m+3); (6m-2); (2m+7) are in A.P.

$\LARGE{\bf{\underline{\underline{TO:FIND:-}}}}$

• The value of m.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Since the terms (8m+3); (6m-2); (2m+7) are in A.P., the common difference will be constant. In other words, if there are three terms a, b and c, and if they are in A.P., then:

$\boxed{\sf{\green{b-a=c-b}}}$

Therefore:

$\bullet \sf (6m-2)-(8m+3) =(2m+7)-(6m-2)$

$\implies \sf 6m-2-8m-3 =2m+7-6m+2$

$\implies \sf -2m-5=-4m+9$

$\implies \sf 2m-5=9$

$\implies \sf 2m=9+5$

$\implies \sf m=\dfrac{14}{2}$

$\implies \boxed{\sf{\green{m=7}}}$

∴The value of m is 7.

SOME IMPORTANT AP FORMULAS:

$\boxed{\substack{\displaystyle \sf a_n=a + (n -1) d \\\\ \displaystyle \sf S = \frac{n}{2} [2a + (n - 1) \times d] \\\\ \displaystyle \sf S = \frac{n}{2} (first \ term + last \ term) \\\\ \displaystyle \sf }}$

Where:

• $\sf a_n$ is the nth term of AP.
• a is the first term of AP.
• n is the number of terms.
• d is the common difference.
• S is the sum of terms.