एका अंकगणिती श्रेढीतील तीन क्रमागत पदांची बेरीज
36 व त्यांचा गुणाकार
1140 आहे, तर ती पदे शोधा.
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Given : एका अंकगणिती श्रेढीतील तीन क्रमागत पदांची बेरीज 36 व त्यांचा गुणाकार
1140 आहे
The sum of three consecutive terms in an arithmetic series
36 and their multiplication 1140, then find those terms.
To Find : पदे
Terms
Solution:
Terms
a- d , a , a + d
a - d + a + a + d = 36
=> 3a = 36
=> a = 12
(12 - d) , 12 , ( 12 + d)
(12 - d)12(12 + d) = 1140
=> 144 - d² = 95
=> d² = 49
=> d = ±7
a = 12 , d = ±7
a- d , a , a + d
=> 5 , 12 , 19
or 19 , 12 , 5
पदे Terms 5 , 12 , 19
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