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in Arithmetic Progression by (31.3k points)
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Find the sum : 

(i) 7 + 10\(\frac{1}{2}\) + 14 + ... + 84 

(ii) 34 + 32 + 30 + ...... + 10 

(iii) 25 + 28 + 31 + .... + 100

(iv) 18 + 15\(\frac{1}{2}\) + 13 + .... + (-49\(\frac{1}{2}\))

1 Answer

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Best answer

(i) a = 7, d =\(\frac{21}{2}\) - 7 = \(\frac{7}{2}\)

Last term, an = 84

an = a + (n – 1)d 

84 = 7 + (n – 1)\(\frac{7}{2}\)

84 = \(\frac{14 + 7n - 7}{2}\)

84 x 2 = 7 + 7n 

168 = 7 + 7n 

7n = 168 – 7 

7n = 161 

n = 23

Hence, sum of given A.P. is \(\frac{2093}{2}\)

(ii)

a = 34, d = 32 – 34 = -2 

Last term, an = 10 

an = a + (n – 1)d 

10 = 34 + (n – 1) (-2) 

10 = 34 – 2n + 2 

2n = 34 – 10 + 2

2n = 26

n = 13 

Sum of n terms,

Hence, Sum of given A.P. is 286

(iii)

a = 25, d = 3 

Last term, an = 100 

An = a + (n -1) d 

100 = 25 (n – 1)3 

75 = 3n – 3 

78 = 3n 

n = 26

sum of n terms,

 

= 13 [50 + 25 x 3] 

= 13 x 125 = 1625

Hence, Sum of given A.P. is 1625

(iv) 

a = 18, d =\(\frac{31}{2}\) -18 = \(\frac{-5}{2}\)

 Last term, an = \(\frac{-99}{2}\)

an = a+ (n – 1)d

\(\frac{-99}{2}\)= 18 + (n – 1)\((\frac{-5}{2})\)

-99 = 36 + 5 – 5n 

-140 = -5n

n = 28

= 7 (36 – 99) 

= 7 (-63) = -441

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