Question 33627
1)given t4 and t7 we need to find out of t5
an arithmetic progression is a sequence starting at a and going up in steps of d 
a, a+d, a+2d,..., a+(n-1)d ... 
here we know 

t4 = a+(4-1)d = a+3d ---- eq 1

t7 = a+(7-1)d = a +6d  --- eq 2

but given t4 = 7 and t7 = 22

so a+3d = 7 and a+6d = 22

solving for a and d

(a+6d)-(a+3d) = 22 - 7 (subtracting both equations)

6d - 3d = 22 - 7 = 15

3d = 15 

d = 5; a = 7 - 3(5) = 7 - 15 = - 8

a = - 8 and d = 5

now to find t5 

t5 = a+(n-1)d
t5 = -8 + (5-1)*5
t5 = -8 +20 = 12

therfore t5 = 12


2)For each arithmetic series,Find S-25
7+13+19+25..... 

Sum of n terms in an A.P is Sn = n/2*{2*a+(n-1)*d}

here a = 7 ;d = 13-7 = 6

S25 = (25/2)*{2*7+(25-1)*6}
S25 = (25/2 )*{14+144}
S25 = (25/2) * 148
S25 = 25*74 = 1850


3)Based on the terms given,state whether or not each sequence is arithmetic, If it is, identify the common difference,d.
15,18,21,24

for a series to be in A.P the common difference should be the same

18 - 15 = 3 ( t2 - t1)

21- 18 = 3  ( t3-t2)

24 - 21 = 3  (t4-t3)

so here the c.d is  same  and hence the series is in A.P and the C.d = 3