Advertisements
Advertisements
प्रश्न
The 4th term of an A.P. is 22, and the 15th term is 66. Find the first term and the common difference. Hence, find the sum of the series to 8 terms.
The 4th term of an A.P. is 22, and the 15th term is 66. Find the sum of its 8 terms.
Advertisements
उत्तर
Let a be the first term and d be the common difference of the given A.P.
Now,
4th term = 22
⇒ a + 3d = 22 ...(i)
15th term = 66
⇒ a + 14d
= 66
Subtracting (i) from (ii), we have
11d = 44
⇒ d = 4
Substituting the value of d in (1), we get
a = 22 − 3 × 4
= 22 − 12
=10
⇒ First term = 10
Now
Sum of 8 terms = `8/2[2xx10+7xx4]`
= 4[20 + 28]
= 4 × 48
= 192
APPEARS IN
संबंधित प्रश्न
Solve the following quadratic equation using formula method only
`3"x"^2 +2 sqrt 5 "x" -5 = 0`
Solve the following quadratic equation using formula method only
`"x"^2 + 1/2 "x" = 3`
ax2 + (4a2 - 3b)x - 12 ab = 0
Find the value of k for which the given equation has real roots:
kx2 - 6x - 2 = 0
Solve the following by reducing them to quadratic equations:
x4 - 26x2 + 25 = 0
Choose the correct answer from the given four options :
If the equation 2x² – 6x + p = 0 has real and different roots, then the values of p are given by
The roots of the quadratic equation 6x2 – x – 2 = 0 are:
(x2 + 1)2 – x2 = 0 has:
The value of k for which the equation x2 + 2(k + 1)x + k2 = 0 has equal roots is:
‘The sum of the ages of a boy and his sister (in years) is 25 and product of their ages is 150. Find their present ages.
