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Question -

Find the sum of the following arithmetic progressions:
(i) 50, 46, 42, тАж. to 10 terms

(ii) 1, 3, 5, 7, тАж to 12 terms

(iii) 3, 9/2, 6, 15/2, тАж to 25 terms

(iv) 41, 36, 31, тАж to 12 terms

(v) a+b, a-b, a-3b, тАж to 22 terms

(vi) (x тАУ y)2, (x2┬а+ y2), (x + y)2,тАж to n terms

(vii) (x тАУ y)/(x + y), (3x тАУ 2y)/(x + y), (5x тАУ 3y)/(x + y), тАж to n terms



Answer -

(i)┬а50, 46, 42, тАж. to 10terms

n = 10

First term, a = a1┬а=50

Common difference, d =a2┬атАУ a1┬а= 46 тАУ 50 = -4

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = 10/2 (100 + (9)(-4))

= 5 (100 тАУ 36)

= 5 (64)

= 320

тИ┤ The sum of the givenAP is 320.

(ii)┬а1, 3, 5, 7, тАж to 12terms

n = 12

First term, a = a1┬а=1

Common difference, d =a2┬атАУ a1┬а= 3 тАУ 1 = 2

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = 12/2 (2(1) +(12-1) (2))

= 6 (2 + (11) (2))

= 6 (2 + 22)

= 6 (24)

= 144

тИ┤ The sum of the givenAP is 144.

(iii)┬а3, 9/2, 6, 15/2, тАж to25 terms

n = 25

First term, a = a1┬а=3

Common difference, d =a2┬атАУ a1┬а= 9/2 тАУ 3 = (9 тАУ 6)/2 = 3/2

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = 25/2 (2(3) +(25-1) (3/2))

= 25/2 (6 + (24)(3/2))

= 25/2 (6 + 36)

= 25/2 (42)

= 25 (21)

= 525

тИ┤ The sum of the givenAP is 525.

(iv)┬а41, 36, 31, тАж to 12terms

n = 12

First term, a = a1┬а=41

Common difference, d =a2┬атАУ a1┬а= 36 тАУ 41 = -5

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = 12/2 (2(41) +(12-1) (-5))

= 6 (82 + (11) (-5))

= 6 (82 тАУ 55)

= 6 (27)

= 162

тИ┤ The sum of the givenAP is 162.

(v)┬аa+b, a-b, a-3b, тАж to22 terms

n = 22

First term, a = a1┬а=a+b

Common difference, d =a2┬атАУ a1┬а= (a-b) тАУ (a+b) = a-b-a-b = -2b

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = 22/2 (2(a+b) +(22-1) (-2b))

= 11 (2a + 2b + (21)(-2b))

= 11 (2a + 2b тАУ 42b)

= 11 (2a тАУ 40b)

= 22a тАУ 440b

тИ┤ The sum of the givenAP is 22a тАУ 440b.

(vi)┬а(x тАУ y)2,(x2┬а+ y2), (x + y)2, тАж to n terms

n = n

First term, a = a1┬а=(x-y)2

Common difference, d =a2┬атАУ a1┬а= (x2┬а+ y2)тАУ (x-y)2┬а= 2xy

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = n/2 (2(x-y)2┬а+(n-1) (2xy))

= n/2 (2 (x2┬а+y2┬атАУ 2xy) + 2xyn тАУ 2xy)

= n/2 ├Ч 2 ((x2┬а+y2┬атАУ 2xy) + xyn тАУ xy)

= n (x2┬а+y2┬атАУ 3xy + xyn)

тИ┤ The sum of the givenAP is n (x2┬а+ y2┬атАУ 3xy + xyn).

(vii)┬а(x тАУ y)/(x + y), (3x тАУ2y)/(x + y), (5x тАУ 3y)/(x + y), тАж to n terms

n = n

First term, a = a1┬а=(x-y)/(x+y)

Common difference, d =a2┬атАУ a1┬а= (3x тАУ 2y)/(x + y) тАУ (x-y)/(x+y) = (2xтАУ y)/(x+y)

By using the formula,

S = n/2 (2a + (n тАУ 1)d)

Substitute the valuesof тАШaтАЩ and тАШdтАЩ, we get

S = n/2(2((x-y)/(x+y)) + (n-1) ((2x тАУ y)/(x+y)))

= n/2(x+y) {n (2x-y) тАУy}

тИ┤ The sum of the givenAP is n/2(x+y) {n (2x-y) тАУ y}

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