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Question

How many terms are there in the series :

(i) 4, 7 , 10, 13, ................ , 148 ?

(ii) 0.5, 0.53, 0.56, .................... , 1.1 ?

(iii) 34,1,114,....................,3?


Solution

(i)common difference (d)

A + (n-1)d = last term
4 + (n-1)3 = 148
      (n-1)3 = 148 - 4
      (n-1)   = 144/3
       n       = 48 + 1
       n       = 49



(ii)common difference =0.03

A + (n-1)d = last term
0.5+(n-1)0.03=1.1
n-1=0.6/0.03
n-1=20
n=21

(iii)c o m m o n space d i f f e r e n c e equals 1 fourth A space plus space left parenthesis n minus 1 right parenthesis d space equals space l a s t space t e r m 3 over 4 plus left parenthesis n minus 1 right parenthesis 1 fourth equals 3 n minus 1 equals open parentheses 9 over 4 close parentheses 4 n minus 1 equals 9 n equals 10

Mathematics
Concise Mathematics
Standard X

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