Very Short Answer on Sequence and Series Class 11 OP Malhotra Exe-14J ISC Maths Solutions Ch-14. In this article you would learn to solve hard questions easily on Sequence and Series. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Sequence and Series Class 11 OP Malhotra Very Short Answer ISC Math Solutions Ch-14
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-14 | Sequence and Series |
| Writer | OP Malhotra |
| Exe-14(J) | Very Short Answer Type Questions. |
Very Short Answer on Sequence and Series
OP Malhotra ISC Class 11 Maths Solutions
Que-1: If an = (−1)n−1 n3, find a9.
Sol: a9 = (−1)8 × 9³
= 1 × 729
= 729
Que-2: Find the value of m if the numbers m + 2, 4m − 6, 3m − 2 are in A.P.
Sol: 2nd term = (1st + 3rd)/2
4m − 6 = [(m+2) + (3m−2)]/2
4m − 6 = 4m/2 = 2m
⇒ 2m = 6
⇒ m = 3
Que-3: Show that 1 + 3 + 5 + …. to n terms = n2.
Sol: Sum of first n odd numbers:
S = 1 + 3 + 5 + … (n terms)
General term of odd numbers = (2n − 1)
S = n/2 × [first term + last term]
= n/2 × [1 + (2n − 1)]
= n/2 × (2n)
= n2
Que-4: For what value of x, the numbers −2/7, x, −7/2 are in G.P.
Sol: For three numbers in G.P., middle term² = product of extremes
x² = (−2/7) × (−7/2)
x² = 1
x = ±1
Que-5: Find the sum of odd integers from 1 to 2001.
Sol: 2001 = 2n − 1
⇒ n = 1001
Sum = n² = 1001²
= 1002001
Que-6: Find the sum of the infinite series 1 + 1/5 + 1/25 + 1/125 + 1/625 + …..
Sol: It is a G.P. with a = 1, r = 1/5
Sum = a/(1−r) = 1/(1−1/5) = 1/(4/5) = 5/4
Que-7: If the 3rd term and the 6th term of an A.P. are 7 and 13 respectively, find the first term and the common difference.
Sol: a + 2d = 7
a + 5d = 13
Subtract ⇒ 3d = 6 ⇒ d = 2
a = 7 − 2(2) = 3
Que-8: Find the first term of the G.P. whose 8th term is 192 and the common ratio is 2.
Sol: T₈ = ar⁷ = 192
a(2⁷) = 192 ⇒ a(128) = 192
a = 192/128 = 3/2
Que-9: If x = 1 + y + y2 + …. to ∞, then find y.
Sol: x = 1/(1 − y)
⇒ 1 − y = 1/x
⇒ y = 1 − 1/x
Que-10: Find the sum of first 8 terms of the geometric series 2 + 6 + 18 + 54 + ….
Sol: a = 2, r = 3, n = 8
S₈ = a(r⁸ − 1)/(r − 1)
= 2(3⁸ − 1)/2
= 6560
Que-11: Two numbers x and y have arithmetic mean 9 and geometric mean 4. Form the quadratic equation whose roots are x and y.
Sol: x + y = 18, xy = 16
Equation: x² − (x+y)x + xy = 0
x² − 18x + 16 = 0
Que-12: A man starts repaying a loan as the first instalment of ₹ 100. If he increases the instalment by ₹ 5 every month, what amount he will pay in the 30th instalment?
Sol: a = 100, d = 5, n = 30
T₃₀ = a + (n−1)d
= 100 + 29×5
= 245
–: End Sequence and Series Class 11 OP Malhotra Exe-14J ISC Maths Ch-14 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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