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प्रश्न
If (2p +1), 13, (5p -3) are in AP, find the value of p.
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उत्तर
Let (2p +1),13,(5p -3) be three consecutive terms of an AP.
Then 13-(2p +1) = (5p -3 )-13
⇒ 7p = 28
⇒ p= 4
∴ When p = 4,(2p +1) ,13 and (5p- 3) from three consecutive terms of an AP .
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संबंधित प्रश्न
In an A.P., if S5 + S7 = 167 and S10=235, then find the A.P., where Sn denotes the sum of its first n terms.
The first and last terms of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there, and what is their sum?
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take `pi = 22/7`)
[Hint: Length of successive semicircles is l1, l2, l3, l4, ... with centres at A, B, A, B, ... respectively.]
Find the sum of all odd natural numbers less than 50.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Is 184 a term of the AP 3, 7, 11, 15, ….?
The 24th term of an AP is twice its 10th term. Show that its 72nd term is 4 times its 15th term.
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
Find the sum of first n terms of an AP whose nth term is (5 - 6n). Hence, find the sum of its first 20 terms.
Draw a triangle PQR in which QR = 6 cm, PQ = 5 cm and times the corresponding sides of ΔPQR?
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
Write the value of a30 − a10 for the A.P. 4, 9, 14, 19, ....
Write the nth term of an A.P. the sum of whose n terms is Sn.
Q.11
If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280. Find the sum of its first n terms.
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
The middle most term(s) of the AP: -11, -7, -3,.... 49 is ______.
