Question

24. If the 5 and 12 terms of an AP are - 4 and - 18 respectively, find the sum of first 20 terms of
the AP
Plss guys solve this question

Answer 1

Step-by-step explanation:

$heloo.... \\ \\ T5 = - 4 \\ a + (n - 1)d = - 4 \\ a + 4d = - 4..............(1) \\ \\ T12 = - 18 \\ a + 11d = - 18...........(2) \\ \\ (2) - (1) = > \\ 7d = - 14 \\ d = - 2 \\ \\ from(1) = > \\ a + 4d = - 4 \\ a + 4( - 2) = - 4 \\ a = 4 \\ \\ S20 = \frac{n}{2} (2a + (n - 1)d) \\ S20 = 10(8 - 38) \\ S20 = - 300 \\ \\ \\ \\ hope \: it \: helps \: uh....$

Answer 2

Answer:

A5=-4

a12=-18

A5=a+(n-1)d

-4=a+(5-1)d

-4=a+4d .....(i)

a12=a+(n-1)d

-18=a+(12-1)d

-18=a+11d. .....(ii)

Divide equation (i) &(ii),we get

Therefore,d=-7/5

Putting value of d in equation (i),we get

-4=a+4d

-4=a+4(-7/5)

-4=a-28/5

-4=5a-28/5

-20=5a-28

-20+28=5a

8/5=a

Sn20=n/2[2a+(n-1)d]

Sn20=20/2[2(8/5)+(20-1)-7/5]

Sn20=10[16/5+19(-7/5)]

Sn20=10[16/5-133/5]

Sn20=10[(16-133)/5]

Sn20=5[-117]

Sn20=-234

Step-by-step explanation: