Question

Find the sum of the following arithmetic progressions:
(i) 50,46,42,... to 10 terms
(ii) 1,3,5,7,... to 12 terms
(iii) 3, $\frac{9}{2},6\frac{15}{2}$, ... to 25 terms
(iv) 41,36,31, ... to 12 terms
(v) a+b,a-b,a-3b,... to 22 terms
(vi) $(x-y{)}^2,(x^2+y^2),(x+y{)}^2$..., to n terms
(vii) $\frac{x-y}{x+y},\frac{3x-2y}{x+y},\frac{5x-3y}{x+y}$,...tn n terms
(viii) -26,-24,-22,... to 36 terms.

Answer 1

Solution:-

In the given problem, we need to find the sum of terms for different A.P.

So, here we use the following formula for the sum of n terms of an A.P.,

Sn=n/2[2a+(n–1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P. n = number of terms

(i) 50, 46, 42,… To 10 terms

Common difference of the A.P. (d)

= a₂ – a₁

= 46 – 50

= -4

Number of terms (n) = 10

First term for the given A.P. (a) = 50

So, using the formula we get,

S10=10/2[2(50)+(10–1)(−4)]

= (5) [ 100 + (9)(-4) ]

= (5) [ 100 – 36 ]

= (6) [64]

= 320

Therefore, the sum of first 10 terms of the given A.P. is 320

(ii) 1, 3, 4, 7, . . . 26 to 12 terms.

Common difference of the A.P. (d)

= a₂ – a₁

= 3 – 1

= 2

Number of terms (n) = 12

First term (a) = 1

So, using the formula we get,

S12=12/2[2(1)+(12–1)(2)]

= (6) [ 2 + (11)(2) ]

= (6) [ 2 + 22 ]

= (6) [24]

= 144

Therefore, the sum of first 10 terms of the given A.P. is 144

(iii) 3,92,6,152,....3,92,6,152,.... to 25 terms.

Common difference here is (d): a₂ – a₁

= 92–392–3

= 9–629–62

= 3232

Number of terms (n) = 25

First term of the A.P. (a) = 3

So, using the formula we get,

S25=25/2[2(3)+(25–1)(32)]

= (25/2)[6+(24)(32)]

= (25/2)[6+(722)]

= (252)[6+36]

= (25/2)[42]

= ( 25 ) ( 21 )

= 525

Therefore, the sum of first 12 terms for the given A.P. is 162.

(iv) 41, 36, 31, . . . To 12 terms.

Common Difference of the A.P. (d) = a₂ – a₁

= 36 – 41

= -5

Number of terms (n) = 12

First term for the given A.P. (a) = 41

So, using the formula we get,

S12=12/2[2(41)+(12–1)(−5)]

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

= (6) [82 – 55]

= (6) [27]

=162

Therefore, the sum of first 12 terms for the given A.P. 162

(v) a + b, a – b, a -3b,… To 22 terms.

Common difference of the A.P. (d) = a₂ – a₁

= (a – b) – (a + b)

=a – b – a – b

= -2b

Number of terms (n) = 22

First term for the given A.P. (a) = a + b

So, using the formula we get,

S22=22/2[2(a+b)+(22–1)(−2b)]

= (11)[ 2(a + b) + (22 – 1)( -2b ) ]

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

= (11)[ 2a + 2b – 42b ]

= (11)[2a – 40b]

= 22a – 40b

Therefore, the sum of first 22 terms for the given A.P. is: 22a – 40b

(vi) (x – y)² , (x², y²), (x + y)², . . . to n terms.

Common difference of the A.P. (d) = a₂ – a₁

= (x² – y²) – (x – y)²

= x² + y² – (x² + y² – 2xy)

= 2xy

Number of terms (n) = n

First term for the given A.P. (a) = (x – y)²

So, using the formula, we get.

Sn=n/2[2(x–y)^2+(n–1)2xy]

Now, taking 2 common from both the terms inside bracket, we get

= n/2(2)[(x–y)^2+(n–1)xy]

= (n)[ ( x – y )² + ( n – 1 )xy ]

Therefore, the sum of first n terms of the given A.P. is (n)[ ( x – y )² + ( n – 1 )xy ]

(vii) (x–y)(x+y),(3x–2y)(x+y),(5x–3y)(x+y),...tonterms(x–y)(x+y),(3x–2y)(x+y),(5x–3y)(x+y),...tonterms

Common difference of the A.P. (d) = a₂ – a₁

= (3x–2yx+y)–(x–yx+y)

= (3x–2y)–(x–y)x+y

= 3x–2y–x+yx+y

= 2x–yx+y2x–yx+y

So, using the formula we get,

Sn=n/2[2(2x–2yx+y)+(n–1)(2x–yx+y)]

= (n2)[(2x–2yx+y)+((n–1)(2x–y)x+y)]

= (n/2)[(2x–2yx+y)+(n(2x–y)–1(2x–y)x+y)]

On further solving, we get

= (n/2)(2x–2y+n(2x–y)−2x+yx+y)

= (n/2)(n(2x–y)–yx+y)

Therefore, the sum of first n terms for the given A.P. is (n/2)(n(2x–y)–yx+y)

(viii) -26, -24, -22, . . . to 36 terms.

Common difference of the A.P. (d) = a₂ – a₁

= (-24) – (-26)

= -24 + 26

=2

Number of terms (n) = 36

First term for the given A.P. (a) = -26

So, using the formula we get,

Sn=(36/2)[2(−26)+(36–1)(2)]

= (18) [ -52 + (35) (2)

= (18) [-52 + 70]

= (18) (18)

= 324

Therefore, the sum of first 36 terms for the given A.P. is 324