NCERT MATHS CLASS - X
Arithmetic Progressions
Exercise 5.3
Question 8:
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Given,
a2 = 14
a3 = 18
d = a3 − a2 = 18 − 14 = 4
a2 = a + d
14 = a + 4
a = 10
= 5610
Question 9:
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
S7 = 49, S17 = 289
7 = (a + 3d)
a + 3d = 7 …………..(i)
Similarly,
17 = (a + 8d)
a + 8d = 17 ………..(ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
= n2
Question 10:
Show that a1, a2 … , an , … form an AP where an is defined as below
(i) an = 3 + 4n
(ii) an = 9 − 5n
Also find the sum of the first 15 terms in each case.
Sol.
(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., an + 1 – an is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
= 15 × 35
= 525
(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., an + 1 – an is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
= −465
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