Question

Consider that 10 arithmetic means are inserted between 3 and 7 and their sum is "a". Again consider that the sum of five numbers in A.P. is 30 and the value of the middle term is "b". Then a + b equals :


1) 16


2) 56


3) 46


4) 36


Please explain your answer :)

Answer 1

Answer:

Option (2) 56

Step-by-step explanation:

Sum of the 10 numbers in AP will be

a = sum of all the 12 numbers in AP - (3 + 7)

= (12/2)(first term + last term) - 10

= 6 × (3 + 7) - 10

= 60 - 10

= 50

Thus, a = 50

If sum of 5 numbers in AP is 30 then let the first of these five numbers be x

using the formula of the sum of an AP

$S=\frac{n}{2}[2x+ (n-1)d]$

or, $30=\frac{5}{2}[2x+(5-1)d]$

or, $12=2x+ 4d$

or, $x+ 2d=6$

But x + 2d gives the thrd term of the five terms in AP    [∵ nthterm of AP is given by a + (n - 1)d]

Therefore the third term = 6

or b = 6

Therefore

a + b = 50 + 6 = 56

Answer 2

Answer:

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

Consider that 10 arithmetic means are inserted between 3 and 7 and their sum is "a". Again consider that the sum of five numbers in A.P. is 30 and the value of the middle term is "b". Then a + b equals :

1) 16

2) 56✅

3) 46

4) 36