find the sum of the first 51 terms of an a.p. whose second and third terms are 14 and 18
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What does a.p mean????
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d=18-14=4
Second term =14
Third term = 18
T2=a+(n-1)*d
14=a+(2-1)*4
14=a+1*4
14=a+4
14-4=a
10=a
S51=n/2(2a+(n-1)d)
=51/2(2*10+(51-1)×4)
=51/2(20+(50*4)
= 51/2(20+200)
=(51/2)*220
=51*110
=5610
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Second term =14
Third term = 18
T2=a+(n-1)*d
14=a+(2-1)*4
14=a+1*4
14=a+4
14-4=a
10=a
S51=n/2(2a+(n-1)d)
=51/2(2*10+(51-1)×4)
=51/2(20+(50*4)
= 51/2(20+200)
=(51/2)*220
=51*110
=5610
HOPE IT HELPS
MATHS ARYABHATA
OMKARA2
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