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ATQ,
A number divisible by both 20 and 30 must be divisible by L.C.M. of 20 and 30 = 60
We look for numbers divisible by 60 in the given range:
First term (a) = 60 × 4 = 240
Common difference (d) = 60
Last term (L) = 60 × 16 = 960
Let the number of terms be n.
Use the A.P. formula:
a + (n - 1)d = L
Substitute values:
240 + (n - 1) × 60 = 960
⇒ 240 + 60n - 60 = 960
⇒ 60n = 780
⇒ n = 780 ÷ 60 = 13
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