Prove that the sum of $n$ number of terms of two different A.P.'s can be same for only one value of $n$ if they have same value of $d, a$ .

Open in App

Solution

Let $n_{1}$ & $n_{2}$ be no. of terms in two diff AP
sum $\frac{n_{1}}{2} [2 a + (n_{1} - 1) d] = \frac{n_{2}}{2} [2 a + (n_{2} - 1) d]$
$2 a n_{1} + n_{1}^{2} d - n_{1} d = 2 a n_{2} + n_{2}^{1} d - n_{2} d$
$\Rightarrow (n_{1} - n_{2}) [2 a + (n_{1} + n_{2}) d - d] = 0$
$\Rightarrow n_{1} = n_{2}$

Suggest Corrections

Similar questions

l

d

s

n

8 d s = (d + 2 l)^{2}

d = 2 a

Question 07

Two atoms are said to be isobars if ?

(a) They have same atomic number but different mass number

(b) They have same number of electrons but different number of neutrons

(d) Sum of the number of protons and neutrons is same but the number of protons is different

If $S_{1}$ be the sum of (2n + 1) terms of an A.P. and $S_{2}$ be the sum of its odd terms, then prove that:

$S_{1} : S_{2} = (2 n + 1) : (n + 1)$

S_{n}

n

S_{12} = 3 (S_{8} - S_{4})

S_{n}

n

S_{30} = 3 (S_{20} - S_{10})

Join BYJU'S Learning Program