Find the sum of the following series (a) 3+7+11+14+(20 terms)
(b)2+6+10+14+(15 terms)(c)30+27+24+21+(40 terms)
(d)5+1+-3+-7+(14 terms)
(e)5+7+8++75
(f)4+7+10++91

need explanation
will give the brainliest!

a) 820

b)450

c)  -540

d) -294

e) 1440

f) 1425

Step-by-step explanation:

a) it is an arithmetic progression with ratio=4

   a1=3,

     a2=3+r=3+4=7

    a20=a1+19r

so a20=3+19*4=3+76=79

S=(a1+an)*n/2 where n=20

so S=(3+79)*20/2=820

b) a15=a1+14*r=2+14*4=2+56=58

S=(a1+a15)*15/2=(2+58)*15/2=60*15/2=30*15=450

It is the same formula for all  the exercises

S=(a1+an)*n/2 ,     where n is the number of terms

c) a40=30+39*(-3)=30-87=-57

So S=(30-57)*40/2=-540

d) a14=5+13*(-4)=5-52=-47

S=(5-47)*14/2=-42*14/2=-42*7=-294

e) it is 5+7+9++75

nr of  terms  = (an-a1) :r+1=(75-5):2+1=36

S=(5+75)*36/2=36*40=1440

f) ratio=7-4=3

the  same formula for n= (91-4):3+1=87:3+1=29+1=30

S=(4+91)*30/2=95*30/2=2850/2=1425

98 points djdjsi zisksis sjsjsi sisksj sksjdj sksks sjsisj

congruent segments

step-by-step explanation: