If , then n = ?
(a)8
(b)7
(c)10
(d)11
Answer with proper explanation.
Question from Class-10 Chapter:– Arithmetic Progressions.
Answers
Correct Question
If , then n = ?
(a)8
(b)7
(c)10
(d)11
Solution -
For n terms:
Sn = n/2 (2a + (n -1)d)
5+9+13+---- (n term)
a = 5, d = 9-5 = 4
Sn = n/2 (2*5 + (n-1)4)
Sn = n/2 (10 + 4n - 4)
Sn = n/2 (4n + 6)
Sn = n(2n + 3) ............(1st equation)
For nth term
S(n+1) = (n+1)/2 (2a + (n+1-1)d)
7+9+11-----(n+1 term)
a = 7, d = 9-7 = 2
S(n+1) = (n+1)/2 (2*7 + (n)2)
S(n+1) = (n+1)/2 (14 + 2n)
S(n+1) = (n+1)(7 + n) .............(2nd equation)
Fraction of both 1st & 2nd equation is 17/16 (given in question)
n(2n+3)/(n+1)(7+n) = 17/16
Cross-multiply them
16n(2n+3) = 17(n+1)(7+n)
32n² + 48n = (17n+17)(7+n)
32n² + 48n = 119n + 17n² + 119 + 17n
32n² + 48n = 136n + 17n² + 119
15n² - 88n - 119 = 0
The above equation is in the form ax² + bx + c = 0
Now, solve it by splitting the middle term.
15n² - 105n + 17n - 119 = 0
15n(n - 7) +17(n - 7) = 0
(n - 7)(15n + 17) = 0
n = +7, -17/15
Negative one neglected. We left with +7.
Therefore, n = 7
Option b) 7
SoluTion:
Correct expression :
- (5 + 9 + 13 + .... n terms)/(7 + 9 + 11 + ....n + 1 terms) = 17/16
We've to find the value of n.
Consider 1st AP : 5 + 9 + 13 + ... n terms
Sum = 17 K
Consider 2nd AP : 7 + 9 + 11 + ...(n + 1) terms
Sum = 16 K
____________
S1 = n/2 [ 2 × 5 + {n - 1) 4]
→ 17K = n (3 + 2n) ........(1)
S2 = (n+1)/2 [ 2 × 7 + 2n ]17
→ 16K = (n + 1)(7 + n) ........(2)
Dividing (1) by (2).
→ [n(3 + 2n) = 17K]/[(n + 1)(7 + n) = 16K]
Cross multiplying it,
→ 17(n + 1)(7 + n) = 16n(3 + 2n)
→ 17(n² + 8n + 7) = 48n + 32n²
→ 17n² + 136n + 119 = 48n + 32n²
→ 17n² - 32n² + 136n - 48n + 119 = 0
→ -15n² + 88n + 119 = 0
Using splitting middle term method, we get,
n = 7 and -17/15
Rejecting the negative value, we get,
→ n = 7
Hence, option B) is correct.