Question

let a1,a2,a3..... be in Ap and h1,h2,h3.... in H.P. If a1=2=H1 and a30=25=h30 then (a7h24+ a14h17) equal to​

Answer 1

Hi,

Answer:

(a7h24+ a14h17) = 100

Step-by-step explanation:

1. We have a1, a2, a3,….. in A.P.  

Also, a1 = 2 & a30 = 25

According to A.P., an = a + (n-1)d

∴ a30 = a + (30-1)d

Or, 25 = 2 + 29d

Or, d = 23/29

∴ a7 = a + (7-1)d = 2 + (6*23/29) = 196/29….. (i)

∴ a14 = a + (14-1)d = 2 + (13*23/29) = 357/29 ….. (ii)

2. We have h1, h2, h3,….. in H.P. when 1/h1, 1/h2, 1/h3,….. are in A.P.

Also, h1 = 2 and h30 = 25

So, 1/h1 = ½ and 1/h30 = 1/25

For A.P., we can also write

1/h30 = 1/h1 + 29D  

Or, 1/25 = ½ + 29D

Or, -23/50 = 29D

Or, D = -23/(50*29)

Now, 1/h24  

= 1/h1 + (24-1) * [-23/(50*29)]  

= ½ + [(23 * -23) / (50*29)]

= ½ - 529/1450

= 196/1450

∴h24 = 1450/196 ……(iii)

Again,  

1/h17

= 1/h1 + (17-1) * [-23/(50*29)]

= ½ + [(16 * -23) / (50*29)]

= ½ - 368/1450

= 357/1450

∴h17 = 1450/357 …. (iv)

3. We have to solve for  

a7h24+ a14h17

substituting the values from eq. (i), (ii), (iii) & (iv)

= [196/29 * 1450/196] + [357/29 * 1450/357]

= 1450/29 + 1450/29

= 50 + 50

= 100