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Gievn that a1 = 2k - 7, a2 = k + 5 and a3 = 3k + 2
d = a2 - a1 = k + 5 - (2k - 7) ............ (i)
d = a3 - a2 = 3k + 2 - (k + 5) ............. (ii)
From equation (i) and (ii)
k + 5 - (2k - 7) = 3k + 2 - (k + 5)
k + 5 - 2k + 7 = 3k + 2 - k - 5
- k + 12 = 2k - 3
2k + k = 12 + 3
3k = 15
k = 15/3
k = 5 Answer
Let the first term = a and common difference = d of the First A.P.
and the first term = a1 and common difference = d1 of the Second A.P.
Here we have to find ratio of 9th term So n will be 2k -1 where k will be 9
Then n = 2k-1 = 2 x 9 - 1 = 17
Now putting the value n = 17 in above equation
Hence Ratio of 9th term = 59:134 Answer
Let the first term of A.P. is a and common difference d = d,
अत: m(am) = n(an)
⇒ m[a + (m – 1)d] = n[a + (n – 1)d]
⇒ am + m(m – 1)d = an + n(n – 1)d
⇒ am – an + [m2 – m – n2 + n]d = 0
⇒ a(m – n ) + [m2 – n2 –( m – n)]d = 0
⇒ a(m – n ) + [(m + n) (m – n) – ( m – n)]d = 0
⇒ (m – n ) [a + {(m + n)– 1}]d = 0
⇒ a + {(m + n)– 1}d = 0 ………….. (1)
am + n = a + {(m + n)– 1}d
am + n = 0 From Equation (i)
Let the first term of A.P. be A and the common difference is d.
ap = a = A + (p -1)d
a = A + (p -1)d …………. (1)
aq = b = A + (q -1)d
b = A + (q -1)d …………. (2)
ar = c = A + (r -1)d
c = A + (r -1)d …………. (3)
a(q – r) + b(r – p) + c(p –q)
= {A + (p -1)d} (q – r) + { A + (q -1)d} (r –p) + { A + (r -1)d}(p – q)
= A{(q – r) + (r – p) + (p – q)} + d {(p – 1) (q – r) + (q – 1) (r –p) + (r -1)(p – q) }
= A (q – r + r – p + p – q) + d (pq – pr – q + r + qr – pq – r + p + pr – qr – p + q)
= A × 0 + d× 0
= 0 Proved
Given that tn = 2n+1
Putting n = 1, 2, 3 respectively;
t1 = 2 x 1 + 1 = 3
t2 = 2 x 2 + 1 = 5
t3 = 2 x 3 + 1 = 7
A.P. : 3, 5, 7 .......... 2n+1 Answer
Given that a = 105, d = 98 - 105 = - 7
Let the nth term is zero
∴ an = a + (n - 1) d
0 = 105 + (n - 1) - 7
0 = 105 - 7n + 7
7n = 112
n = 112/7
n = 16 Answer
Given that : a = 114, d = 109 - 114 = - 5
Let the first negative term n
∴ an = - 1
an = a + (n - 1) d
- 1 = 114 + (n - 1) - 5
- 1 = 114 - 5n + 5
5n = 114 + 5 + 1
5n = 120
n = 120/5
n = 24 Answer
Hence 24th term is the frist negative
Important Question for board and preboard exams
Given that : Sn = 3n2 + 4
Now Putting n = 1
S1 = 3(1)2 + 4
= 3 + 4
= 7
Putting n = 2
S2 = 3(2)2 + 4
= 3 x 4 + 4
= 12 + 4
= 16
a1 ≠ S1 ≠ 7
∵ a1 = S1 - S0 Here S0 ≠ 0
S0 = 3n2 + 4 ⇒ 3(0)2 + 4 ⇒ 0 + 4 = 4
Using an = Sn - S(n - 1)
a1 = S1 - S0
= 7 - 4
= 3
a2 = S2 - S1 [ using formulla an = Sn - S(n - 1) ]
= 16 - 7
= 9
d = a2 - a1 ⇒ 9 - 3 = 6
an = a + (n -1)d
= 3 + (n - 1) 6
= 3 + 6n - 6
= 6n - 3
Hence nth term is 6n - 3 Answer
Second Method :
Given that : Sn = 3n2 + 4 ................ (i)
Replace n by (n - 1)
We have : Sn - 1 = 3(n - 1)2 + 4
⇒ Sn - 1 = 3(n2 - 2n + 1) + 4
= 3n2 - 6n + 3 + 4
= 3n2 - 6n + 7 ......................... (ii)
an = Sn - S(n - 1)
= 3n2 + 4 - (3n2 - 6n + 7 )
= 3n2 + 4 - 3n2 + 6n - 7
= 6n - 3 Answer
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