Question -
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}