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4 November, 14:45

An AP has 15 terms and a common

difference of - 3. Find its first and last terms

if its sum is

b 15

d - 120

a 120

+1
Answers (1)
  1. 4 November, 15:05
    0
    Step-by-step explanation:

    Sum = n/2{2a + (n-1) d

    When sum=15 and d=-3

    15=15/2{2a + (15-1) - 3}

    15=7.5{2a + (14) - 3}

    15=7.5{2a + (-42) }

    15=7.5 (2a-42)

    15=15a-315

    15+315=15a

    330=15a

    a=330/15

    =22

    a=22

    Last term=a + (n-1) d

    =22 + (15-1) - 3

    =22 + (14) - 3

    =22-42

    = - 20

    When sum=120

    Sum=n/2{2a + (n-1) d

    120=15/2{2a + (15-1}-3

    120=7.5{2a + (14) - 3}

    120=7.5{2a + (-42) }

    120=7.5 (2a-42)

    120=15a-315

    120+315=15a

    435=15a

    a=435/15

    =29

    a=29

    Last term=a + (n-1) d

    =29 + (15-1) - 3

    =29 + (14) - 3

    =29-42

    = - 13

    When sum = - 120

    Sum=n/2{2a + (n-1) d

    -120=15/2{2a + (15-1) - 3

    -120=7.5{2a + (14) - 3}

    -120=7.5{2a + (-42) }

    -120=7.5 (2a-42)

    -120=15a-315

    -120+315=15a

    195=15a

    a=195/15

    a=13

    Last term=a + (n-1) d

    =13 + (15-1) - 3

    =13-42

    = - 29
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