Question

# In an AP (i) given a=5,d=3,an=50, find n and Sn. (ii) given a12=37,d=3, find a and S12. (iii) given d=5,S9=75, find a and a9. (iv) given a=8,an=62,Sn=210, find n and d. (v) given l=28,S=144, and there are total of 9 terms. Find a.

Solution

## (i) Given a=5, d=3, an=50, find n and Sn. Using formula an=a+ (n−1) d,   to find  nth term of arithmetic progression, we can say that ⇒an=5+ (n−1)(3) ⇒50=5+3n−3 ⇒48=3n ⇒n=483=16 Applying formula, Sn=n2(2a+ (n−1)d) to find sum of n terms of AP, we get S16=16/2(10+ (16−1) (3)) =8(10+45) =8(55) =440 Therefore, n=16 and Sn=440 (ii) Given a12=37, d=3, find a and S12. Using formula an=a+ (n−1) d,   to find nth term of arithmetic progression, we can say that a12=a + (12−1) 3 ⇒37=a+33 ⇒a=4 Applying formula, Sn=n2(2a + (n−1) d) to find sum of n terms of AP, we get S12=122(8+ (12−1) (3) =6(8+33) =6(41) =246 Therefore, a=4 and S12=246 (iii) Given d=5, S9=75, find a and a9. Applying formula, Sn=n2 (2a+ (n−1) d) to find sum of n terms of AP, we get S9=92 (2a+ (9−1) 5) ⇒75=92 (2a+40) ⇒150=18a+360 ⇒−210=18a ⇒a=−21018 =−353  Using formula an=a+ (n−1) d,   to find nth term of arithmetic progression, we can say that a9=−353 + (9−1) (5) =−353 +40 = (−35+120)3 = 853  Therefore, a=−353 and a9=853 (iv) Given a=8, an=62, Sn=210, find n and d. Using formula an=a+ (n−1) d,   to find nth term of arithmetic progression, we can say that 62=8+ (n−1) (d) =8+nd−d ⇒62=8+nd−d ⇒nd−d=54 ⇒nd=54+d                    … (1) Applying formula, Sn=n2 (2a+ (n−1) d) to find sum of n terms of AP, we get 210=n2 (16+ (n−1) d) =n2 (16+nd−d) Putting equation (1) in the above equation, we get 210=n2 (16+54+d−d) =n2 (70) ⇒21070 =n2  ⇒n=6 Putting value of n in equation (1), we get ⇒6d=54+d ⇒d=545  Therefore, n=6 and d=545  (v) Given l=28, S=144, and there are total of 9 terms. Find a. Applying formula, Sn=n2 [a +l], to find sum of n terms, we get 144=92 [a+28] ⇒2889 =[a+28] ⇒32=a+28 ⇒a=4 Mathematics

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