WebJun 25, 2024 · Nth term of an A.P. ⇒ aₙ = 3n - 2. As we know that, Put the value of n = 1 in the equation, we get. ⇒ 3(1) - 2. ⇒ 3 - 2 = 1. Put the value of n = 2 in the equation, we get. … WebDec 26, 2024 · The fourth term of AP whose nth term is 3n + 2 is 14. Given : The AP whose nth term is 3n + 2. To find : The fourth term of the AP . Solution : Step 1 of 2 : Write down the given data . Here it is given that for the AP nth term is 3n + 2 . Step 2 of 2 : Find fourth term of AP . Putting n = 4 in the given AP we get . ⇒ Fourth term = ( 3 × 4 ) + 2
The nth term of a sequence is 3n - 2. Is the sequence an A.P. ? If so
WebJan 19, 2024 · The `n^ (th)` term of the sequence is `3n-2` . Is the sequence an AP. If so; find the 10th term . Doubtnut 2.64M subscribers 8.7K views 3 years ago The `n^ (th)` term of … WebThe nth term is given by 3n+2, so for the first term, this is when n=1, so we substitute 1 as n and do a similar thing for 2nd and 3rd terms. 1st term (n=1): 3 (1)+2= 3+2=5 2nd term (n=2): 3 (2)+2=6+2=8 3rd term (n=3): 3 (3)+2= 9+2=11 ... the 10th term of the sequence is when n=10 so we substitute n=10 into the nth term formula which gives: 3 … thicknesser planer for sale south africa
If the sum of n terms of an A.P. be 3 n2 n and its common
WebMar 22, 2024 · Ex 9.2,13 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m. Let a1, a2, … an be the given A.P Given, Sum of n terms = 3n2 + 5n Sn = 3n2 + 5n Putting n = 1 in (1) S1 = 3 × 12 + 5 × 1 = 3 × 1 + 5 × 1 = 3 + 5 = 8 Sum of first 1 terms = First term ∴ First term = a1 = S1 = 8 Sn = 3n2 + 5n … WebApr 11, 2024 · To find the N th term in the Arithmetic Progression series we use the simple formula . T N = a 1 + (N-1) * d Recommended: Please try your approach on {IDE} first, … WebMar 30, 2024 · Transcript. Example 6 The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 12th terms. There are 2 AP s with different first term and common difference For the first AP Let first term be a common difference be d Sum of n term = Sn = /2 (2a + (n 1)d) & nth term = an = a + (n 1)d ... sail bonus share