Question

M1,M2, M3, M4, M5, M6, M7
In the given sequence, each term is equal
to succeeding minus the constant d. If
M2+ M4+ M6= 48, what is the sum of
M1,M3,M5,M7?

Answer 1

Answer:

M1+M3+M5+M7= 64

Step-by-step explanation:

To find the sum of M1,M3,M5,M7

Firstly we have to identify the sequence,each term is equal

to succeeding minus the constant d,so the given sequence is an AP.

M1= a

M2=a+d

M3=a+2d

.

.

.

M7= a+6d

ATQ

M2+ M4+ M6= 48

a+d+a+3d+a+5d=48

3a+9d=48

a+3d=16... eq1

To find M1+M3+M5+M7

= a+a+2d+a+4d+a+6d

=4a+12d

=4(a+3d)

put the value of a+3d from eq1

= 4(16)

= 64

So,M1+M3+M5+M7= 64

Hope it helps you.