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ANSWER 3.
n = 28
tn = 4n – 3
t1 = 4 x 1 – 3 = 1
t2 = 4 x 2 – 3 = 5
t28 = 4 x 28 – 3
= 112 – 3 = 109
∴ a = 1, d = t2 – t1 = 5 – 1 = 4
L = 109.
Sn = (n/2) (2a + (n – 1)d)
S28 = (28/2) (2 x 1 + 27 x 4)
= 14(2 + 108)
= 14 x 110
= 1540
Answer 4.
Let tn be nth term of an A.P.
tn = Sn – Sn - 1
= 2n2 – 3n – [2(n – 1)2 – 3(n – 1)] = 2n2 – 3n – [2(n2 – 2n + 1) – 3n + 3]
= 2n2 – 3n – [2n2 – 4n + 2 – 3n + 3]
= 2n2 – 3n – [2n2 – 7n + 5]
= 2n2 – 3n – 2n2 + 7n – 5
tn = 4n – 5
t1 = 4(1) – 5 = 4 – 5 = -1
t2 = 4(2) -5 = 8 – 5 = 3
t3 = 4(3) – 5 = 12 – 5 = 7
t4 = 4(4) – 5 = 16 – 5 = 11
The A.P. is -1, 3, 7, 11,… The common difference is 4
Ans. 5
given a 104= 125
a4=0
now.
to find- S35.
a + 103d=125
a+3d = 0
subtracting we get.
100d =125
d=125/100=5/4
now. d=5/4
a= -15/4..
S35= 35/2(2×-15/4 +34×5/4)
=35/2(-15/2+85/2)
= 35/2×35
=1225/2=612.5
therefore sum of first 35 terms is 612.5.
Ans 6.
Sum of odd positive integer less than 450
1 + 3 + 5 + …. 449
Here a = 1,
d = 3 – 1 = 2,
l = 449
Aliter: Sum of all the positive odd intergers = n2
= 225 × 225 = 50625
Sum of the odd integers less than 450 = 50625
Ans.7
The required sum
=(sum of all natural numbers from 603 to 901)-(sum of all natural numbers between 602 and 902 which are divisible by 4)
=224848-56400
=168448
The required sum
=(sum of all natural numbers from 603 to 901)-(sum of all natural numbers between 602 and 902 which are divisible by 4)
=224848-56400
=168448