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Arithmetic

Progressions

  • Additional sums based on concepts of AP

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Q.1] The 24th term of an AP is twice its 10th term. Show that its

72nd term is 4 times its 15th term.

Sol:

Show that

a72 =

4 (a15)

Lets check, what is given?

a24 =

2 (a10)

a24 = a + 23d

a + 23d =

2

(a + 9d)

∴

a10 = a + 9d

a + 23d =

2a

+ 18d

∴

23d – 18d =

2a – a

∴

5d =

a

∴

….(i)

Lets simplify LHS & RHS

LHS,

a72

= a + 71d

= 5d + 71d

= 76d

RHS,

4(a15)

= 4(a + 14d)

= 4(5d + 14d)

= 4(29d)

= 76d

∴

a72 = 4 (a15)

a72 = a + 71d

a15 = a + 14d

Additional example

HOMEWORK

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Q.2] The sum of first n terms of an AP is 5n2 + 3n. If its mth term is 168, find the value of m. Also, find the 20th term of this AP.

Sol:

Sn =

5n2 + 3n

Find S1 by putting n = 1

∴ S1 =

5(1)2

+ 3(1)

= 5

+ 3

= 8

S1 means sum of first term i.e. first term itself

∴ S1 =

a1 =

8

Find S2 by putting n = 2

a2 = S2 – S1

5(2)2

+ 3(2)

= 5(4)

+ 6

= 20

+ 6

= 26

– S1

= 26 – 8

= 18

∴ S2 =

d =

a2 – a1

= 18 – 8

= 10

am

= 168

am = a + (m – 1)d

∴ a + (m – 1)d

= 168

∴ 8

+ (m – 1)

10

= 168

∴ (m – 1)10

= 160

∴ m – 1

= 16

∴ m

= 17

a20

= a + 19d

= 8

+ 19(10)

= 8 + 190

= 198

a20

Additional example