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
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
If (k – 3), (2k + l) and (4k + 3) are three consecutive terms of an A.P., find the value of k.
In the following determine the set of values of k for which the given quadratic equation has real roots:
2x2 + 3x + k = 0
In the quadratic equation kx2 − 6x − 1 = 0, determine the values of k for which the equation does not have any real root.
Determine, if 3 is a root of the given equation
`sqrt(x^2 - 4x + 3) + sqrt(x^2 - 9) = sqrt(4x^2 - 14x + 16)`.
If x = 2 and x = 3 are roots of the equation 3x² – 2kx + 2m = 0. Find the values of k and m.
Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: px2 – 4x + 3 = 0
If the difference of the roots of the equation x2 – bx + c = 0 is 1, then:
Find the nature of the roots of the quadratic equation:
4x2 – 5x – 1 = 0
Find the value of ‘c’ for which the quadratic equation
(c + 1) x2 - 6(c + 1) x + 3(c + 9) = 0; c ≠ - 1
has real and equal roots.
