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## The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.

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Sol.      Let ‘a’ be the first term and ‘d’ the common difference.

Given : t4 = –15
For t4 put n = 4
∴        a + (4 – 1)d = –15
∴                a + 3d = –15                  ...(I)
t9 = a + (9 – 1)d
For t9 put n = 9
∴               a + 8d = –30                   ...(ii)
Subtracting eq. (I) from (II),

a   +   8d   =   –30
–     a   +   3d   =   –15
(–)   (–)           (+)

5d   =   –15
∴                  d   =   –3
Substituting d = –3, in eqn (I),
a + 3 × (–3)  =  15
a – 9    = –15
a = –15 + 9
∴                        a = –6
∴     a = –6 and d = –3
Sum of first 10 terms,

Sn${n\over2}[2a+(n-1)d]$

For S10 put n = 10
S10 =  $10\over 2$ [ 2(–6) + (10 – 1)(–3)]
S10 =  5[–12 + 9 (–3)]
S10 =  5 [–12 –27]
S10 = –195
∴    Sum of first ten numbers =  –195

∴    Sum of first ten numbers = –195

Arithmetic Progression September 28 , 2018 0 Comments 102 views