# In an A.P., show that a_m+n + a_mâ€“n = 2a_m.

$We\;know\;the\;first\;term\;is\;a\;and\;the\;common\;difference\\*of\;an\;A.P\;is\;d.\\*Given:\;a_{m+n}+a_{m-n}=2a_{m}\\* By\;using\;the\;formula, \\* \Rightarrow a_{n}=a+(n-1)d\\* Now,\;let\;us\;take\;LHS,\\* \Rightarrow a_{m+n}+a_{m-n}\\*\Rightarrow a_{m+n}+a_{m-n}=a+(m+n-1)d+a+(m-n-1)d\\*\Rightarrow a+md+nd-d+a+md-nd-d\\*\Rightarrow 2a+2md-2d\\* \Rightarrow 2(a+md-d)\\*\Rightarrow\;2[a+d(m-1)]\;\;\;\;\;\;\;\;\;\;\;{\therefore a_{n}=a+(n-1)d}\\*\Rightarrow a_{m+n}+a_{m-n}=2a_{m}\\* Hence\;Proved.$

## Most Viewed Questions

### Preparation Products

##### Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
##### Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
##### NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
##### NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
##### NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Boost your Preparation for JEE Main with our Foundation Course

Exams
Articles
Questions