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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > If 7th and 13th terms of an AP be 34 and 64 respectively....
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If 7th and 13th terms of an AP be 34 and 64 respectively. Then its 18th term is :?

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Let 1st term be "a" and common difference be "d" then 7 term and 13 term can be written as a+6d=34......equ.(1) a+12d=64......equ.(2) Solve both equ. We get a=4 & d=5 then So, 18 term be a+17d= 4+17×5 = 89 And.
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T7 = 34 where n = 7 and let the first term is a and common difference is d , a + (n-1)*d = 34 a + (7-1)*d = 34 a + 6d =34 ................................................................................................(1) T13 = 64 where n = 13 a + (n-1) * d = 64 a + (13-1) *d = 64 a + 12d = 64 ............................................................................................(2) Now...
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T7=34 where n=7 and let the first term is a and common difference is d , a + (n-1)*d = 34 a + (7-1)*d = 34 a + 6d=34 ................................................................................................(1) T13=64 where n=13 a + (n-1) * d=64 a + (13-1) *d = 64 a + 12d=64 ............................................................................................(2) Now solving equation (1) and (2) 2(a + 6d=34 ) a + 12d=64 __________________________ a=4 ________________________ putting a=4 in equation (1) 4 + 6d=34 => 6d = 34-4 d = 30/6 = 5 T18 = a + (18-1) * d T18 = 4 + 17 * 5 T18 = 4 + 85 = 89 read less
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18th term is 89
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a+6d=34 -(i) a+12d=64 -(ii) Solve both these equations , u will get d=5 & putting the value of d u will get a=4. Now for 18th term =89 .
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for 7th term, n=7an=24putting the formula,an=a (n-2)dso,24=a 6d...............eq1 now for 13th term,n=13an=64puttin above formula we get,64=a 12d.....eq 2eqating 1
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for 7th term,  n=7 an=24 putting the formula,an=a+(n-2)d so,24=a+6d...............eq1   now for 13th term, n=13 an=64 puttin above formula we get,64=a+12d.....eq 2 eqating 1&2 a+6d=24 a+12d=64 now we get, a+6d-a-12d=24-64 -6d=-40 d=40/6=20/3(putting value of d=20/3 in eq 1,value ofa=-16 now...
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for 7th term,  n=7 an=24 putting the formula,an=a+(n-2)d so,24=a+6d...............eq1   now for 13th term, n=13 an=64 puttin above formula we get,64=a+12d.....eq 2 eqating 1&2 a+6d=24 a+12d=64 now we get, a+6d-a-12d=24-64 -6d=-40 d=40/6=20/3(putting value of d=20/3 in eq 1,value ofa=-16 now for 18th term n=18, a=-16 a18=-16+17(20/3) =-48/3=340/3 =292/3 read less
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RAJESH SIR, IIT-JEE TEACHER

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