Question

Q.35
Q.32
Sum of the first 14 terms of an A.P. is 1505
and its first term is 10. Find its 25th term.
In the given figure
triangle the length
10 cm and ADBC
BD=8 cm. Find the
(Take V3 = 1.732
(1) 370
(3) 380
(2) 320
(4) 390
Q.33 With vertices A, B and C of a triangle ABC as
centres, arcs are drawn with radii 5 cm each
as shown in figure.
B
(1) 19.3 cm?
(3) 17.3 cm
IfAB = 14 cm, BC = 48 cm and CA= 50 cm,
then find the area of the shaded region.
"
(Use =310)
Q.36 Out of the 120 p​

Answer 1

Answer:

Step-by-step explanation:

N/2×(2a+(n-1)d)=sn

14/2(20+(14-1)d)=1505

=>140+91d=1505

=>91d=1365

=>d=15/7

A+(n-1)d=an

10+(25-1)15/7=an

An=430/7

Answer 2

Answer:

370

Step-by-step explanation:

S14=1505

a=10

a25=a+24d

Sn=n/2(2a+(n-1)d)

S14=14/2(2(10)+(14-1)d)

=1505=7(20+13d)

13d=195

d=15

a25=a+24d

=10+24(15)

=370