Answer:
1) given, a10=52
a+9d=52 ...........(1)
also, a17=20+a13
a+16d=20+a+12d
16d−12d=20
4d=20
d=5
putting the value of d in eq. (1), we get,
a+9(5)=52
a+45=52
a=7
hence, the required AP is 7,12,17,.........
2)504 , 511 , 518 , ....,889, 896 are multiples of 7 between 500 and 900 and these are in Arithmetic progression .
First term = a = 504,
Common difference = a2 - a1
d = 511 - 504 = 7
Let the number of terms = n
Last term ( L ) = 896
We know that ,
a + ( n - 1 )d = L
504 + ( n - 1 ) 7 = 896
Divide each term with 7 , we get
72 + n - 1 = 128
71 + n = 128
n = 128 - 71
n = 57
Therefore, Sum of n terms ( Sn )= n/2 [ a + L ]
Sum of the 57 terms = S57
S57 = (57/2 )[ 504 + 896 ]
=( 57/2 ) [ 1400 ]
= 57 × 700 = 39900
Sum of 7 multiples between 500 and 900 = 39900
3)Let 1st term of A.P=a and common difference =d
We have to prove that
S12=3(S8−S4)
S12=12/2[2a+(12−1)d]
S8= 8/2[2a+(8−1)d]
S4=4/2[2a+(4−1)d]
For the given A.P.,
Now, R.H.S= 3(S8−S4) = 3(8/2[2a+(8−1)d]-4/2[2a+(4−1)d])
=3[8x+28d−4a−6d]
=6(2a+11d)
=12/2[2a+(12−1)d]
=S12
R.H.S= L.H.S
4)First term(a)=3
Last term(tn)=253
Common Difference(d)=8−3=5
∵tn=a+(n−1)d
⇒253=3+(n−1)×5
⇒253−3=(n−1)×5
⇒250÷5=n−1
⇒50+1=n
∴n=51
In 51 terms, 20th term from the last term will be 51−30+1=22th
∴t22=a+(22−1)d
=3+21×5=3+105
=108
5)Given sequence is 3,15,27,39,...
The first term is a=3
The common difference is d=15−3=12
we know that the nth term of the arithmetic progression is given by a+(n−1)d
Let the mth term be 120 more than the 21st term
Therefore, 120+mthterm=21stterm
⟹120+a+(m−1)d=a+(21−1)d
⟹120+(m−1)12=(20)12
⟹12m−12=240+120
⟹12m=360+12
⟹12m=372
⟹m=372/12 = 31
Therefore, the 31st term is 120 more than the 21st term
Hope this helps you :)
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Verified answer
Answer:
1) given, a10=52
a+9d=52 ...........(1)
also, a17=20+a13
a+16d=20+a+12d
16d−12d=20
4d=20
d=5
putting the value of d in eq. (1), we get,
a+9(5)=52
a+45=52
a=7
hence, the required AP is 7,12,17,.........
2)504 , 511 , 518 , ....,889, 896 are multiples of 7 between 500 and 900 and these are in Arithmetic progression .
First term = a = 504,
Common difference = a2 - a1
d = 511 - 504 = 7
Let the number of terms = n
Last term ( L ) = 896
We know that ,
a + ( n - 1 )d = L
504 + ( n - 1 ) 7 = 896
Divide each term with 7 , we get
72 + n - 1 = 128
71 + n = 128
n = 128 - 71
n = 57
Therefore, Sum of n terms ( Sn )= n/2 [ a + L ]
Sum of the 57 terms = S57
S57 = (57/2 )[ 504 + 896 ]
=( 57/2 ) [ 1400 ]
= 57 × 700 = 39900
Sum of 7 multiples between 500 and 900 = 39900
3)Let 1st term of A.P=a and common difference =d
We have to prove that
S12=3(S8−S4)
S12=12/2[2a+(12−1)d]
S8= 8/2[2a+(8−1)d]
S4=4/2[2a+(4−1)d]
For the given A.P.,
Now, R.H.S= 3(S8−S4) = 3(8/2[2a+(8−1)d]-4/2[2a+(4−1)d])
=3[8x+28d−4a−6d]
=6(2a+11d)
=12/2[2a+(12−1)d]
=S12
R.H.S= L.H.S
4)First term(a)=3
Last term(tn)=253
Common Difference(d)=8−3=5
∵tn=a+(n−1)d
⇒253=3+(n−1)×5
⇒253−3=(n−1)×5
⇒250÷5=n−1
⇒50+1=n
∴n=51
In 51 terms, 20th term from the last term will be 51−30+1=22th
∴t22=a+(22−1)d
=3+21×5=3+105
=108
5)Given sequence is 3,15,27,39,...
The first term is a=3
The common difference is d=15−3=12
we know that the nth term of the arithmetic progression is given by a+(n−1)d
Let the mth term be 120 more than the 21st term
Therefore, 120+mthterm=21stterm
⟹120+a+(m−1)d=a+(21−1)d
⟹120+(m−1)12=(20)12
⟹12m−12=240+120
⟹12m=360+12
⟹12m=372
⟹m=372/12 = 31
Therefore, the 31st term is 120 more than the 21st term
Hope this helps you :)