a1 a2 a3 are in AP if a1+a2+ a3 is equal to 27 and a1 square + a2 square + a3 square is equal to 275 find a1 a2 a3
Answers
Answer:
Try solving according to this:
a
6
is
41
11
.
Step-by-step explanation:
Given:
\frac{a_1+a_2+a_3+...+a_p}{a_1+a_2+a_3+...+a_q}=\frac{p^2}{q^2}
a
1
+a
2
+a
3
+...+a
q
a
1
+a
2
+a
3
+...+a
p
=
q
2
p
2
we know that,
S_n=\frac{n}{2}(2a+(n-1)d)S
n
=
2
n
(2a+(n−1)d)
So,
\frac{\frac{p}{2}(2a+(p-1)d)}{\frac{q}{2}(2a+(q-1)d)}=\frac{p^2}{q^2}
2
q
(2a+(q−1)d)
2
p
(2a+(p−1)d)
=
q
2
p
2
\frac{2a+(p-1)d}{2a+(q-1)d}=\frac{p}{q}
2a+(q−1)d
2a+(p−1)d
=
q
p
\frac{2a+pd-d}{2a+qd-d}=\frac{p}{q}
2a+qd−d
2a+pd−d
=
q
p
(2a+pd-d)q=(2a+qd-d)p(2a+pd−d)q=(2a+qd−d)p
2aq+pqd-qd=2ap+pqd-pd2aq+pqd−qd=2ap+pqd−pd
2aq-2ap=-pqd+qd+pqd-pd2aq−2ap=−pqd+qd+pqd−pd
2a(q-p)=(q-p)d2a(q−p)=(q−p)d
2a=d2a=d
Now, using a_n=a+(n-1)da
n
=a+(n−1)d
\frac{a_6}{a_{21}}=\frac{a+(6-1)d}{a+(21-1)d}
a
21
a
6
=
a+(21−1)d
a+(6−1)d
\frac{a_6}{a_{21}}=\frac{a+5(2a)}{a+20(2a)}
a
21
a
6
=
a+20(2a)
a+5(2a)
\frac{a_6}{a_{21}}=\frac{a+10a}{a+40a}
a
21
a
6
=
a+40a
a+10a
\frac{a_6}{a_{21}}=\frac{11a}{41a}
a
21
a
6
=
41a
11a
\frac{a_6}{a_{21}}=\frac{11}{41}
a
21
a
6
=
41
11
Therefore, Value of \frac{a_6}{a_{21}}\:is\:\frac{11}{41}
a
21
a
6
is
41
11
Answer:
a1+a2+a3=27 (1)
a1^2+a2^2+a3^2=275 (2)
Let assume that
a1=a-d
a2=a
a3=a+d
.
.
.
.
Put these assumptions in eq 1
(a-d)+a+(a+d) =27
a-d+a+a+d=27
3a=27
a=9 (3)
Now put the above assumption in eq 2
(a-d)^2+a^2+(a+d)^2=275
a^2-2ad+d^2+a^2+a^2+2ad+d^2=275
3a^2+2d^2=275
Put the of a from eq3
3(9)^2+2d^2=275
243+2d^2=275
2d^2=275-243
2d^2=32
d^2=32/2=16.
d=4. (4)
Put the value of a and d to find a1, a2, a3
a1=a-d=9-4=5
a2=a=9
a3=a+d=9+4=13
So the sequence is
5,9,13,.........