{"id":4494,"date":"2021-08-20T05:25:17","date_gmt":"2021-08-20T05:25:17","guid":{"rendered":"https:\/\/mathemerize.com\/?p=4494"},"modified":"2021-11-17T17:47:23","modified_gmt":"2021-11-17T12:17:23","slug":"formula-for-arithmetic-progression","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/formula-for-arithmetic-progression\/","title":{"rendered":"Formula for Arithmetic Progression (AP)"},"content":{"rendered":"
Here you you will learn what is arithmetic progression (AP) and formula for arithmetic progression.<\/p>\n
Let’s begin –<\/p>\n
A sequence is called an arithmetic progression if the difference of a term and the previous term is always same\u00a0 i.e.\u00a0<\/p>\n
\n\\(a_{n+1}\\) – \\(a_n\\) = constant (=d) for all n \\(\\in\\) N<\/p>\n
The constant difference generally denoted by d is called the common difference.<\/p>\n<\/blockquote>\n
for example<\/span> : 1, 4, 7, 10 ….. is an AP whose first term is 1 and the common difference is equal to 4 – 1 = 3.<\/p>\n
Also Read<\/strong> : Formula for Geometric Progression (GP)<\/a><\/p>\n
Formula for Arithmetic Progression<\/h2>\n
(a) General term of an AP ( nth term of ap)<\/h3>\n
Let a be the first term and d be the common difference of an AP. Then its nth term or general term is a + (n – 1)d\u00a0<\/p>\n
\ni.e.\u00a0 \u00a0\\(a_n\\) = a + (n – 1)d.<\/p>\n<\/blockquote>\n
(b) nth term of an AP from the end<\/h3>\n
Let a be the first term and d be the common difference of an AP having m terms. Then nth term from the end is \\((m – n + 1)^{th}\\) term from the beginning.<\/p>\n
\n\\(\\therefore\\)\u00a0 nth term from the end\u00a0 = \\(a_{m-n+1}\\)<\/p>\n
= a + (m-n+1-1)d = a + (m-n)d<\/p>\n<\/blockquote>\n
Also nth term from the end = \\(a_m\\) + (n-1)(-d)<\/p>\n
[\\(\\because\\)\u00a0 \u00a0Taking \\(a_m\\) as the first term and the common difference equal to ‘-d’ ]<\/p>\n
(c) Sum to n terms of an AP<\/h3>\n
The sum \\(S_n\\) of n terms of an AP with first term ‘a’ and common difference ‘d’ is<\/p>\n
\n\\(S_n\\) = \\(n\\over 2\\) [2a + (n-1)d]<\/p>\n
or,\u00a0 \\(S_n\\) = \\(n\\over 2\\) [a + l] ,\u00a0 where l = last term = a + (n-1)d<\/p>\n<\/blockquote>\n\n\n
Example : <\/span>Show that the sequence 9, 12, 15, 18, ……. is an AP. find its 16th term, general term sum of first 20 terms.<\/p>\n
Solution : <\/span>We have, (12 – 9) = (15 – 12) = (18 – 15) = 3. Therefore, the given sequence is an AP with the common difference 3.
\nfirst term = a = 9
\n\\(\\therefore\\) 16th term = \\(a_{16}\\) = a + (16-1)d
\n\\(\\implies\\) \\(a_{16}\\) = 9 + 15*3 = 54
\n\\(\\because\\) General term = nth term = \\(a_n\\) = a + (n-1)d
\n\\(\\therefore\\) \\(a_n\\) = 9 + (n-1)*3 = 3n + 6
\nNow, sum of first 20 terms = \\(S_{20}\\) = \\(20\\over 2\\) [2*9 + (20-1)3]
\n\\(S_{20}\\) = 10[18 + 19*3]
\n= 750
<\/p>\n\n\n
\nRelated Questions<\/h3>\n