A.P. // Arithmetic Progression// Objective Questions & Answers// Class 10h //समांतर श्रेणी//
1. If an A.P. is the common term of 3n + 5, then its common difference will be:
यदि किसी A.P. का सामान्य पद 3n + 5 है तो इसका सार्व अंतर होगा:
Explain:- Put difference value of n = 0 or 1 or 2 or 3 etc in 3n+5
Then ,
a1=3n+5=3x0+5=5
a2=3n+5=3x1+5=8
a3=3n+5=3x2+5=11
a4=3n+5=3x3+5=14
Hence, A.P. will be 5,8,11,14.........
Now , d= a2-a1 = 8-5 = 3
2. The common difference of A.P. 14,20,,26,22,28.....will be :
A.P. 4,20,,26,22,28.....का सार्व अंतर होगा :
3. When the first term of an AP is 3 and the common difference is 3, then the AP has 3 terms:
जब एपी के प्रथम पद 3 तथा सार्व अंतर 3 है तो ए पी के 3 पद होंगे:
4. Which term of the sequence 5, 7, 9, 11 is 27?
अनुक्रम 5,7, 9, 11 का कौन - सा पद 27 है?
5. If the 6th and twelfth term of an A.P. is 13 and 25 respectively, then its 20th term will be:
यदि किसी A.P. का छठा और बारहवां पद क्रमशः 13 और 25 है, तो इसका 20वां पद होगा:
6. The nth term of an A.P. is given by an = 3 + 4n. The common difference is
किसी समांतर श्रेणी का nवाँ पद an = 3 + 4n द्वारा दिया जाता है। सामान्य अंतर है
Explain:- We have an = 3 + 4n ∴ an+1 = 3 + 4(n + 1) = 7 + 4n ∴ d = an+1 – an = (7 + 4n) – (3 + 4n) = 7 – 3 = 4
7. If p, q, r and s are in A.P. then r – q is
यदि p, q, r और s समांतर श्रेणी में हैं तो r - q है
Explain:- Since p, q, r, s are in A.P. ∴ (q – p) = (r – q) = (s – r) = d (common difference)
8. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
यदि एक एपी में तीन संख्याओं का योग 9 है और उनका गुणनफल 24 है, तो संख्याएं हैं
Explain:- Let three numbers be a – d, a, a + d
∴ a – d +a + a + d = 9
⇒ 3a = 9
⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2
9. The (n – 1)th term of an A.P. is given by 7,12,17, 22,… is
किसी समांतर श्रेढ़ी का (n – 1)वाँ पद 7,12,17, 22,… द्वारा दिया जाता है
Explain:- Here a = 7, d = 12-7 = 5
∴ an-1 = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5n – 3
10. The nth term of an A.P. 5, 2, -1, -4, -7 … is
एपी 5, 2, -1, -4, -7 ... का nवां पद है
Explain:- Here a = 5, d = 2 – 5 = -3, an = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n
11. The 11th term from the end of the A.P. -5, -11, -15,…, -1100 is
ए.पी. -5, -11, -15,…, -1100 के अंत से 11वाँ पद है
Explain:- Here al = -1100, d = -11 – (-5) = -11 + 5 = – 5
∴ 11th term from the end = a – (n – 1 )d = -1100 – (11 – 1) (-5) = -1100 + 45 = -955
12. Find the sum of 12 terms of an A.P. whose nth term is given by an = 3n + 4
उस समांतर श्रेणी के 12 पदों का योग ज्ञात कीजिए जिसका nवाँ पद an = 3n + 4 . द्वारा दिया गया है
Explain:- Here an = 3n + 4
∴ a1 = 7,
a2 = 10,
a3 = 13
∴ a= 7, d = 10 – 7 = 3
∴ S12 = 12/2[2 × 7 + (12 – 1) ×3]
= 6[14 + 33]
= 6 × 47 = 282
13. The sum of all two digit odd numbers is
दो अंकों की सभी विषम संख्याओं का योग होता है
Explain:- All two digit odd numbers are 11,13,15,… 99, which are in A.P. Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = 45/2[11 + 99]
= 45/2 × 110
= 45 × 55 = 2475
14. The sum of first n odd natural numbers is
प्रथम n विषम प्राकृत संख्याओं का योग है
Explain:- Required Sum = 1 + 3 + 5 + … + upto n terms.
Here a = 1, d = 3 – 1 = 2
Sum = n/2[2 × 1 + (n – 1) × 2]
= n/2[2 + 2n – 2]
= n/2 × 2n = n²
15. If (p + q)th term of an A.P. is m and (p – q)th term is n, then pth term is
यदि किसी समांतर श्रेणी का (p + q)वाँ पद m है और (p - q)वाँ पद n है, तो pवाँ पद है
Explain:- Let a is first term and d is common difference
∴ ap + q = m
ap – q = n
⇒ a + (p + q – 1)d = m = …(i)
⇒ a + (p – q – 1)d = n = …(ii)
On adding (i) and (ii), we get
2a + (2p – 2)d = m + n
⇒ a + (p -1)d = m+n/2 …[Dividing by 2
∴ ap = m+n/2
16. If a, b, c are in A.P. then a−b/b−c is equal to
यदि a, b, c एपी में हैं तो a−b/b−c बराबर . है
Explain:- Since a, b, c are in A.P.
∴ b – a = c – b
⇒ b−a/c−b = 1
⇒ a−b/b−c = 1
17. If a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is
यदि a, b, c, d, e एपी में हैं, तो a - 4b + 6c - 4d + e का मान है
Explain:- Let common difference of A.P. be x
∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x
Given equation a-4b + 6c-4d + e
= a – 4(a + x) + 6(A + 2x) – 4(a + 3x) + (a + 4x)
= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x
= 8a – 8a + 17x – 17x
= 0
18. The 10th term from the end of the A.P. 4, 9,14, …, 254 is
एपी 4, 9,14, …, 254 के अंत से 10वां पद है
Explain:- Here an = 254, d = 9-4 = 5 ∴ 10th term from the end = 254 – (10 – 1 )5 = 254 = 9(5) = 254 – 45 = 209
19. If 2x, x + 10, 3x + 2 are in A.P., then x is equal to
यदि 2x, x + 10, 3x + 2 समांतर श्रेणी में हैं, तो x बराबर है
Explain:- Since 2x, x + 10 and 3x + 2 are in A.P.
∴ 2(x + 10) = 2x + (3x + 2) {a1+a3 = 2a2}
⇒ 2x + 20 = 5x + 2
⇒ 2x – 5x = 2 – 20
⇒ 3x = 19 ⇒ x = 6
20. The sum of all odd integers between 2 and 100 divisible by 3 is
2 और 100 के बीच 3 से विभाज्य सभी विषम पूर्णांकों का योग है
Explain:- The numbers are 3, 9,15, 21, …, 99
Here a = 3, d = 6 and an = 99
∴ an = a + (n – 1 )d
⇒ 99 = 3 + (n – 1) x 6
⇒ 99 = 3 + 6n – 6
⇒ 6n = 102
⇒ n = 17
Required Sum = n/2[a + an]
= 17/2[3 + 99]
= 17/2 × 102
= 867
21. If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, then 18th term is
यदि किसी समांतर श्रेणी के 7वें पद का 7 गुना उसके 11वें पद के 11 गुना के बराबर है, तो 18वां पद है
Explain:- We have 7a7 = 11a11
⇒ 7(a + 6d) = 11(a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 4a = -68d
⇒ a = -17d
∴ a18 = a + 17d
= -17d + 17d = 0
22. If p, q, r are in AP, then p3 + r3 – 8q3 is equal to
यदि p, q, r AP में हैं, तो p3 + r3 - 8q3 बराबर है
Explain:- ∵ p, q, r are in AP.
∴ 2q = p + r
⇒ p + r – 2q = 0
∴ p3 + r3 + (-2p)3 = 3 × p × r × -2q [Using if a + b + c = 0 ⇒ a3 + b3 + c3 = 3 abc]
⇒ p3 + r3 – 8q3 = -6pqr.
23. In an AP, if a = 3.5, d = 0, n = 101, then a will be
एक AP में, यदि a = 3.5, d = 0, n = 101, तो an होगा
Explain:- a101 = 3.5 + 0(100) = 3.5
24. The list of numbers -10, -6, -2, 2, … is.
संख्याओं की सूची -10, -6, -2, 2,… है।
Explain:- No need
25. Two APs have the same common difference. . The first term of one of these is -1 and that of the other is -8. Then the difference between their 4th terms is
दो एपी में समान सामान्य अंतर है। . इनमें से एक का प्रथम पद -1 है और दूसरे का -8 है। तो उनके चौथे पदों के बीच का अंतर है
Explain:- a4 – b4 = (a1 + 3d) – (b1 + 3d) = a1 – b1= – 1 – (-8) = 7
26. In an AP, if d = -2, n = 5 and an = 0, the value of a is
एक AP में, यदि d = -2, n = 5 और an = 0, तो a का मान होता है
Explain:- d = – 2, n = 5, an = 0
∵ an = 0
⇒ a + (n – 1)d=0
⇒ a + (5 – 1)(- 2) = 0
⇒ a = 8
Correct option is (d).
27. If the common difference of an AP is 3, then a20 – a15 is
यदि किसी AP का सार्व अंतर 3 है, तो a20 - a15 है
Explain:- Common difference, d = 3
a20 – a15
= (a + 19d) – (a+ 14d)
= 5d
=5 × 3
= 15
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28. The next term of the AP √18, √50, √98, …….. is
AP √18, √50, 98, …….. का अगला पद है
Explain:- √18, √50, √98, …..
= 3√2, 5√2, 7√2, ……
∴ Next term is 9√2 = √162
29. The common difference of the AP 1/p , 1-p/p , 1-2p/p......is
AP 1/p , 1-p/p , 1-2p/p ...... का सामान्य अंतर
Explain:- Common difference = a2 – a1
30. If the nth term of an AP is (2n +1), then the sum of its first three terms is
यदि किसी AP का nवाँ पद (2n +1) है, तो इसके प्रथम तीन पदों का योग है
Explain:- a1= 2 × 1 + 1 = 3,
a2 = 2 × 2 + 1 = 5,
a3 = 2 × 3 + l= 7
∴ Sum = 3 + 5 + 7 = 15
