In an ap sum of first n terms is 3n 2/2
WebMar 3, 2024 · Given that Sum to n terms of an AP is . 3n² + n. Given common difference is 6, If suppose a is the first term of A.P and 'd' is the . common difference, the Sum to n terms of an A.P is. given by Sn = (n/2)*[2a + (n - 1)d] Given Sn = 3n² + n. If we substitute n = 1 , in Sn we get Sum to 1 tems of an A.P . which is equivalent to first term of A ... WebThe Sum of the First N Term Sofa an Ap is `( (3n^2) /2 +(5n) /2)`. Find Its Nth Term and the 25th Term . CBSE English Medium Class 10. Question Papers 939. Textbook Solutions 33590. MCQ Online Mock Tests 12. Important Solutions 4010. Question Bank Solutions 26519. Concept Notes & Videos 213.
In an ap sum of first n terms is 3n 2/2
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WebLet S n denotes the sum of first n terms of the AP. ∴ s n = 3n 2 + 5n. ⇒ `s_(n-1) = 3 (n-1) ^2 + 5 (n-1)` = `3(n^2 - 2n + 1) + 5 (n-1)` =`3n^2 -n-2` Now , n th term of AP , a n = s n - s n-1 = (3n 2 + 5n ) - ( 3n 2-n-2) = 6n + 2 . Let d be the common difference of the AP. ∴ d = a n - a n-1 = (6n + 2 ) - [ 6(n-1 ) +2] = 6n + 2 - 6 (n-1) -2 = 6 WebMar 30, 2024 · 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 Similarly for second AP Let first term = A common difference = D Sn = /2 (2A + (n 1)D) & nth term = An = A + (n 1)D We need to find ratio of 12th term i.e. …
WebWhat is the value of x in the equation 2^(x+1) - 3(2^x) + 2 = 0? If the sum of the first n terms of an arithmetic sequence is given by S = 3n^2 - 2n, what is the common difference of the sequence? what is the degree of a polynomial 5x⁴+6x²-8x-9 WebIn an A.P the sum of first n terms is 3n2/2 + 13n/2. Find the 25th term arithmetic progression cbse class-10 1 Answer +1 vote answered Sep 30, 2024 by KajalAgarwal …
WebAug 28, 2024 · Let the sum of n terms be given by Sn so Sn = 3n²/2+ 5n/2 S1 = 3 (1)²/2 + 5 (1)/2 = 3/2+5/2 => 4 so 1st term is 4 say 'a' Now S2 = 3 (2)²/2 + 5 (2)/2 = 6+5 => 11 Now a2 …
WebLet the sum of n terms be given by SnsoSn = 3n²/2+5n/2S1 = 3(1)²/2+5(1)/2= 3/2+5/2= 4So 1st term is 4 say aNowS2 = 3(2)²/2+5(2)/2= 6+5= 11Now a2 =S2−a1=> a2 =11−4= 7Now …
WebSep 15, 2024 · Solution: We can use our concept of AP here to solve the problem as follows: Step 1: Calculate the first number divisible by 3 in the given range: 20 = 3 × 6 + 2 a = 20 + (3 - remainder) a = 20 + (3 - 2) a = 21. Step 2: Calculate the Last number (nth) in the given range:-. greeley plumbing alexandria mnWebMar 27, 2024 · Solution For 10] Find the sum of first ' n ' terms of an AP 2 ,8 ,18 ,32 , 11] If α,β and γ are the zeroes of the polynomial P(x)=x3−6x2−x+30 then fin. The world’s only live instant tutoring platform. Become a tutor About us Student login Tutor login. Login. Student Tutor. Filo instant Ask button for chrome browser. ... greeley planning codeWebAn arithmetic progression or arithmetic sequence (AP) is a sequence of numbers such that the difference from any succeeding term to its preceding term remains constant throughout the sequence. The constant difference is called common difference of that arithmetic progression. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression … greeley playsWebIn an AP, the sum of first n terms is 3n^22 + 13n2 Find the 25th term. Class 11 >> Applied Mathematics >> Sequences and series >> Arithmetic progression >> In an AP, the sum of … greeley plumbers who take paymentsWebApr 13, 2024 · If the sum of the first n terms of an arithmetic sequence is 3n^2 + n, and the first term is 1, what is the common difference of the sequence? Toggle navigation. Home; Ask A Question; Answer Questions; Users; Support/FAQ; Register; Login; Math Problem. yaspal2510. Question. If the sum of the first n terms of an arithmetic sequence is 3n^2 + … flower grass pngWebApr 8, 2024 · Let the sum of n terms be given by Sn. so. Sn = 3n²/2+ 5n/2. S1 = 3(1)²/2 + 5(1)/2 = 3/2+5/2 => 4. so 1st term is 4 say 'a' Now. S2 = 3(2)²/2 + 5(2)/2 = 6+5 => 11. Now … flower grass clipartWebAnswer (1 of 4): Given that, Sn = 3n^2–4n S1 = 3–4=(-1 ) S2 = 12–8=4 S1 = a = (-1) S2 = a1+a2 = a+a+d =2a+d=4 = d =4–2a = 4–2(-1) =4+2=6 So, a = -1 & d = 6 ... greeley plumbing and heating reynoldsville